Chapter 9 · Ray Optics and Optical Instruments

9.4 Refraction through a Prism and Optical Instruments

A prism turns light through an angle that is smallest when the ray passes symmetrically — and that minimum gives the refractive index. Lenses in the right places make microscopes that enlarge the very small and telescopes that bring the very distant close.

In 1609 Galileo heard of a Dutch spectacle-maker's "spyglass", built one himself from two lenses in a tube, and pointed it at the sky. Within months he had seen the mountains of the Moon, the phases of Venus and four moons circling Jupiter. Half a century later Robert Hooke and Antonie van Leeuwenhoek turned lenses on a drop of pond water and found it full of living cells. Two simple lens arrangements — the telescope and the microscope — opened the very large and the very small to human eyes, and their modern descendants range from the Giant Magellan Telescope, now being built in Chile, to the microscope on a school laboratory bench.

This section finishes the chapter. First we follow a ray through a glass prism, find the angle through which it is deviated, and show that the deviation has a minimum value from which the refractive index of the glass can be measured very accurately. Then we look at the eye, at the difference between seeing an object large and seeing it close, and at the magnifying power of the simple microscope, the compound microscope and the astronomical telescope.

Refraction through a Prism

A prism has two plane refracting faces meeting at an edge; the angle $A$ between them is the angle of the prism. A ray in the plane perpendicular to the edge meets the first face at angle of incidence $i$, refracts to $r_1$, crosses the prism, meets the second face at $r_2$ and emerges at the angle of emergence $e$. It has been turned towards the base of the prism through the angle of deviation $\delta$.

The normals to the two faces meet at a point $N$, and in the quadrilateral formed by the edge, the two points of incidence and $N$, two angles are right angles. So the angle between the normals is $180^\circ-A$, and in the triangle formed by the ray inside the prism and the two normals, $r_1+r_2+(180^\circ-A)=180^\circ$. The deviation at the first face is $i-r_1$ and at the second $e-r_2$, both towards the base; adding them gives the total.

📐 Result 9.12 — Refraction through a prism

For a ray passing through a prism of angle $A$ in a principal section, $$r_1+r_2=A,\qquad \delta=i+e-A,$$ with $\sin i=n\sin r_1$ and $\sin e=n\sin r_2$ for a prism of refractive index $n$ in air. A ray can emerge from the second face only if $r_2$ does not exceed the critical angle $C$ (Section 9.2).

Plot $\delta$ against $i$ (Interactive 1): the deviation is large for small and for very large angles of incidence, and passes through a single minimum, the angle of minimum deviation $\delta_m$. Because light paths are reversible, if a ray entering at $i$ leaves at $e$, a ray entering at $e$ leaves at $i$ with the same deviation. The minimum can occur at only one angle, so it must be where $i=e$. Then $r_1=r_2=r$, and $r_1+r_2=A$ gives $r=A/2$; also $\delta_m=2i-A$, so $i=\dfrac{A+\delta_m}{2}$. At minimum deviation the ray passes through the prism symmetrically, parallel to the base of an isosceles prism.

📐 Result 9.13 — Refractive index from the angle of minimum deviation

At minimum deviation $i=e$ and $r_1=r_2=A/2$, so $$n=\frac{\sin\left(\dfrac{A+\delta_m}{2}\right)}{\sin\left(\dfrac A2\right)}.$$ Measuring $A$ and $\delta_m$ with a spectrometer gives $n$ to four or five significant figures — the standard laboratory method.

📐 Result 9.14 — Deviation by a thin prism

For a prism of small angle $A$ (a few degrees) and small angles of incidence, $\sin\theta\approx\theta$ gives $i\approx nr_1$ and $e\approx nr_2$, so $$\delta=(n-1)A,$$ independent of the angle of incidence (as long as it is small).

✏️ Example 1 — A ray through a 60° prism
A ray strikes one face of an equilateral glass prism ($A=60^\circ$, $n=1.50$) at $i=40^\circ$. (a) Find $r_1$, $r_2$, the angle of emergence and the deviation. (b) Find the angle of minimum deviation and the angle of incidence that gives it.
(a) First face $\sin r_1=\dfrac{\sin40^\circ}{1.50}=0.4285$, $r_1\approx25.4^\circ$, so $r_2=60^\circ-25.4^\circ=34.6^\circ$ (below $C=41.8^\circ$, so the ray emerges).
Second face $\sin e=1.50\sin34.6^\circ=0.852$, $e\approx58.5^\circ$; $\delta=40^\circ+58.5^\circ-60^\circ\approx38.5^\circ$.
(b) Minimum $r=30^\circ$: $\sin i=1.50\sin30^\circ=0.75$, $i\approx48.6^\circ$, and $\delta_m=2i-A\approx37.2^\circ$ — slightly less than in (a), as it must be.
Check Interactive 1 ($A=60^\circ$, $n=1.50$) at $i=40^\circ$ reads $e=58.5^\circ$, $\delta=38.5^\circ$, and its curve has its minimum $37.2^\circ$ at $i=48.6^\circ$. ✓
✏️ Example 2 — Measuring a refractive index, and a thin prism
(a) A student finds the angle of minimum deviation of an equilateral prism to be $38.0^\circ$. Find the refractive index of the glass. (b) A thin prism of the same glass has an angle of $4.0^\circ$. Through what angle does it deviate a ray?
(a) $n=\dfrac{\sin49.0^\circ}{\sin30^\circ}=\dfrac{0.7547}{0.5}\approx1.51$.
(b) $\delta=(n-1)A=0.51\times4.0^\circ\approx2.0^\circ$, for any small angle of incidence.
Interactive 1 — Deviation by a Prism and the Minimum

The ray is traced through the prism with Snell's law at each face. Change the angle of incidence (or press Play): the dot runs along the graph of deviation $\delta$ against $i$. At the bottom of the curve the ray inside the prism is parallel to the base and $i=e$; the readout then recovers $n$ from $\delta_m$ with Result 9.13. For small $i$ the ray is totally reflected at the second face and does not emerge.

40°
$r_1$, $r_2$—
emergence $e$—
deviation $\delta$—
$\delta_m$ and $n$ from it—

Colours. The refractive index of glass is slightly larger for violet light than for red, so a prism deviates violet light a little more and spreads white light into a spectrum. You met this splitting (and the rainbow and the scattering of sunlight that are related to it) in Class 10; it is not part of this chapter's syllabus. In this chapter $n$ means the index for one colour, usually yellow light.

The Eye and the Simple Microscope

The eye (in brief). Light enters the eye through the cornea, which does most of the bending, and the eye lens behind it fine-tunes the focus so that a real, inverted image forms on the retina, about $2$ cm behind. To see near and far objects the ciliary muscles change the curvature, and hence the power, of the eye lens — accommodation. A normal young eye sees clearly from infinity (its far point) down to about $25$ cm (its near point); a short-sighted eye cannot focus distant objects and is corrected with a diverging lens, a long-sighted eye cannot focus near objects and is corrected with a converging lens (Class 10).

How large an object looks depends on the angle it subtends at the eye, which fixes the size of its image on the retina. Bringing an object closer increases that angle — but only down to the near point, closer than which it blurs. An optical instrument helps by forming an image that subtends a larger angle while still lying where the eye can focus on it.

📘 Definition 9.7 — Near point and magnifying power

The near point is the closest point at which the eye can focus clearly; for a normal eye it is at the least distance of distinct vision, $D=25$ cm. The magnifying power (angular magnification) $m$ of an optical instrument is the ratio of the angle $\beta$ subtended at the eye by the final image to the angle $\alpha$ subtended by the object seen directly by the unaided eye — at the near point $D$ for a microscope or magnifier, or at its actual (very large) distance for a telescope: $$m=\frac{\beta}{\alpha}.$$

The simple microscope. A converging lens of short focal length $f$, held close to the eye with a small object of height $h$ just inside its focus, forms a virtual, erect, enlarged image. If the image is at the near point, $v=-D$, the linear magnification is $m=\dfrac vu=1-\dfrac vf=1+\dfrac Df$; since the image is at $D$, the ratio of angles is the same. If instead the object is placed at the focus, the image is at infinity — more relaxed for the eye — and the image subtends $\beta=h/f$ while the object at the near point would subtend $\alpha=h/D$.

📐 Result 9.15 — Magnifying power of a simple microscope

For a converging lens of focal length $f$ held close to the eye, $$m=1+\frac Df\quad\text{(image at the near point)},$$ $$m=\frac Df\quad\text{(image at infinity)},$$ with $D=25$ cm. A lens of focal length $5$ cm magnifies $6$ or $5$ times. Lenses with focal lengths below about $1$ cm blur the image badly, so a single lens is limited to roughly $\times10$.

✏️ Example 3 — A watchmaker's magnifier
A watchmaker uses a magnifying glass of focal length $5.0$ cm. (a) Find its magnifying power when the image is at the near point, and where the object must then be. (b) Find its magnifying power with the image at infinity. (c) A screw $1.0$ mm long is examined in setting (b). What angle does its image subtend at the eye?
(a) $m=1+\dfrac{25}{5.0}=6.0$. Cartesian convention, $v=-25$ cm, $f=+5.0$ cm: $\dfrac1u=\dfrac1v-\dfrac1f=-\dfrac1{25}-\dfrac15=-\dfrac{6}{25}$, $u\approx-4.2$ cm.
(b) $m=\dfrac{25}{5.0}=5.0$, with the screw at the focus, $5.0$ cm from the lens.
(c) $\beta=\dfrac hf=\dfrac{1.0\ \text{mm}}{50\ \text{mm}}=0.020$ rad ($\approx1.1^\circ$), five times the $0.0040$ rad it subtends at $25$ cm.

The Compound Microscope and Telescopes

The compound microscope. To magnify more, use two converging lenses. The objective, of very short focal length $f_o$, faces the object, which is placed just beyond its focus; it forms a real, inverted, enlarged image $A'B'$ inside the tube. The eyepiece, of short focal length $f_e$, acts as a simple microscope for viewing that image. The magnifying power is the product of the objective's linear magnification and the eyepiece's angular magnification, $m=m_om_e$. The objective's image lies near the eyepiece's focus; if $L$, the tube length, is the distance between the second focus of the objective and the first focus of the eyepiece, similar triangles give $m_o\approx-\dfrac{L}{f_o}$ (the image is about $L$ beyond the objective's focus).

📐 Result 9.16 — Magnifying power of a compound microscope

With the objective's linear magnification $m_o=\dfrac{v_o}{u_o}$, $$m=m_om_e,$$ with $m_e=1+\dfrac{D}{f_e}$ (final image at the near point) or $m_e=\dfrac D{f_e}$ (final image at infinity). For the final image at infinity, approximately $$|m|\approx\frac{L}{f_o}\cdot\frac{D}{f_e}.$$ The final image is inverted. Short focal lengths for both lenses give high magnifying power.

✏️ Example 4 — Inside a laboratory microscope
A compound microscope has an objective of focal length $1.0$ cm and an eyepiece of focal length $2.5$ cm. An object is placed $1.2$ cm in front of the objective. (a) Find the position of the intermediate image and the objective's magnification. (b) Find the magnifying power and the separation of the lenses when the final image is at infinity. (c) Repeat (b) for the final image at the near point.
(a) Objective Cartesian convention: $u_o=-1.2$ cm, $f_o=+1.0$ cm. $\dfrac1{v_o}=1.0-\dfrac1{1.2}=\dfrac16$, $v_o=+6.0$ cm; $m_o=\dfrac{6.0}{-1.2}=-5.0$.
(b) Infinity The intermediate image must be at the eyepiece's focus: separation $=6.0+2.5=8.5$ cm. $m_e=\dfrac{25}{2.5}=10$, so $m=-5.0\times10=-50$: magnifying power $50$, image inverted.
(c) Near point $m_e=1+\dfrac{25}{2.5}=11$, $m=-55$. For the eyepiece, $v_e=-25$ cm: $\dfrac1{u_e}=-\dfrac1{25}-\dfrac1{2.5}=-\dfrac{11}{25}$, $u_e\approx-2.27$ cm, so the separation is $6.0+2.27\approx8.3$ cm.
Check Interactive 2 (microscope, object at $1.20$ cm) reads $m_o=-5.00$, $m_e=10.0$, magnifying power $50$, separation $8.50$ cm. ✓

The refracting (astronomical) telescope. Here the object is very far away, so it cannot be brought closer; the aim is to increase the angle it subtends. The objective, of long focal length $f_o$ and large aperture, forms a real, inverted image of the distant object in its focal plane. The eyepiece, of short focal length $f_e$, views this image; in normal adjustment the image lies at the eyepiece's focus and the final image is at infinity. A distant object subtending a small angle $\alpha$ forms an image of height $h=f_o\alpha$; seen through the eyepiece it subtends $\beta=h/f_e$. Hence $m=\beta/\alpha=f_o/f_e$.

📐 Result 9.17 — Magnifying power of an astronomical telescope

In normal adjustment (final image at infinity), $$m=\frac{f_o}{f_e},$$ and the length of the tube is $f_o+f_e$. If the final image is formed at the near point instead, $m=\dfrac{f_o}{f_e}\left(1+\dfrac{f_e}{D}\right)$. The final image is inverted — no drawback for astronomy.

A large objective also collects more light, so fainter objects can be seen. Very large lenses, however, are heavy, can be supported only at their edges, and bring different colours to slightly different foci. All modern large telescopes are therefore reflecting telescopes: the objective is a concave (usually paraboloidal) mirror, which has no colour error, can be supported over its whole back and can be made in segments. In the Cassegrain design a small convex secondary mirror sends the converging light back through a hole in the centre of the primary to an eyepiece or camera behind it, folding a long focal length into a short tube. The $3.6$ m Devasthal Optical Telescope in Uttarakhand and the $10$ m Keck telescopes in Hawaii are reflectors of this family.

✏️ Example 5 — Looking at the Moon
An astronomical telescope has an objective of focal length $1.0$ m and an eyepiece of focal length $4.0$ cm. The Moon subtends about $0.0090$ rad at the Earth. (a) Find the magnifying power in normal adjustment and the length of the telescope. (b) Find the diameter of the Moon's image formed by the objective. (c) What angle does the final image subtend at the eye?
(a) $m=\dfrac{100}{4.0}=25$; length $=100+4.0=104$ cm.
(b) $h=f_o\alpha=100\times0.0090=0.90$ cm.
(c) $\beta=m\alpha=25\times0.0090=0.225$ rad $\approx13^\circ$ — the Moon looks as wide as a hand held at arm's length.
Check Interactive 2 (telescope, $f_e=4.0$ cm) reads $m=25$ for any small $\alpha$.
✏️ Example 6 — Choosing lenses
A student has three converging lenses of powers $+1.0$ D, $+10$ D and $+40$ D. (a) Which two make the telescope with the largest magnifying power, and what are its magnifying power and length in normal adjustment? (b) Which two would you choose for a compound microscope, and which would be the objective?
Focal lengths $100$ cm, $10$ cm and $2.5$ cm.
(a) Telescope Longest $f_o$, shortest $f_e$: objective $+1.0$ D, eyepiece $+40$ D. $m=\dfrac{100}{2.5}=40$, length $102.5$ cm.
(b) Microscope Both focal lengths should be short: the $+40$ D lens ($2.5$ cm) as objective and the $+10$ D lens ($10$ cm) as eyepiece. Lesson: a telescope's objective has the longer focal length; a microscope's has the shorter.
Interactive 2 — Ray Tracer for a Microscope and a Telescope

Both instruments are in normal adjustment: the objective's image (red arrow) lies at the eyepiece's focus, so the rays leave the eyepiece parallel and the eye sees the final image at infinity. Microscope ($f_o=1.0$ cm, $f_e=2.5$ cm): move the object towards the objective's focus and watch the intermediate image grow. Telescope ($f_o=100$ cm): a parallel beam from a distant star at angle $\alpha$ leaves the eyepiece at the larger angle $\beta$. The readouts are computed from the lens formula; the vertical scale is stretched. Press Play to sweep the slider.

1.20 cm
objective mₒ—
eyepiece mₑ—
magnifying power—
lens separation—
⚠️ Common mistakes with prisms and instruments

1. Using $n=\dfrac{\sin i}{\sin(A/2)}$ away from minimum deviation. $r=A/2$ only at minimum deviation; elsewhere find $r_1$ and use $r_2=A-r_1$.

2. Degrees in $\delta=(n-1)A$. The thin-prism formula works with $A$ and $\delta$ in the same unit (degrees or radians) — but only for small angles.

3. Mixing up magnification and magnifying power. Instruments are rated by the ratio of angles. A telescope's image is far smaller than the Moon, but it subtends a larger angle.

4. Swapping the lenses. Telescope: long $f_o$, short $f_e$, $m=f_o/f_e$. Microscope: both short, with $f_o\lt f_e$ usually.

5. Forgetting which adjustment. "Normal adjustment" or "relaxed eye" means final image at infinity ($m_e=D/f_e$); "image at the near point" or "least distance of distinct vision" means $m_e=1+D/f_e$.

🚀 Beyond the board syllabus (JEE Advanced) — dispersion by a prism

In JEE Advanced 2026 only (it lists deviation and dispersion of light by a prism); not in the CBSE, JEE Main or NEET 2026 syllabi, which list only refraction through a prism.

Because $n$ depends on colour, a thin prism deviates violet light through $\delta_v=(n_v-1)A$ and red light through $\delta_r=(n_r-1)A$. The angular dispersion is the angle between them, $$\delta_v-\delta_r=(n_v-n_r)A,$$ and the dispersive power of the glass is the ratio of the angular dispersion to the mean deviation $\delta_y=(n_y-1)A$ (yellow light), $$\omega=\frac{n_v-n_r}{n_y-1}.$$ With typical (illustrative) values for crown glass, $n_r=1.514$, $n_y=1.517$, $n_v=1.528$, a $6.0^\circ$ prism deviates yellow light by $3.10^\circ$ and spreads the spectrum over $0.084^\circ$; $\omega\approx0.027$. Two thin prisms of different glasses, placed with their angles opposed, can cancel the dispersion while leaving a net deviation (an achromatic combination) or cancel the deviation while leaving dispersion (a direct-vision prism). Syllabus note: dispersion by a prism is in the JEE Advanced syllabus; it is not in the CBSE, NEET or JEE Main syllabus.

Exercises

I
The prism.
  • 1 A ray meets one face of an equilateral prism ($n=1.50$) at $50^\circ$. Find $r_1$, $r_2$, $e$ and $\delta$.
  • 2 The angle of minimum deviation of a $60^\circ$ prism is $30^\circ$. Find its refractive index.
  • 3 A $60^\circ$ prism has refractive index $\sqrt3$. Find its angle of minimum deviation.
  • 4 Find the deviation produced by a thin prism of angle $5.0^\circ$ and refractive index $1.6$.
  • 5 A ray falls normally on one face of a prism of angle $30^\circ$ ($n=1.5$). Find the angle of emergence and the deviation.
  • 6 At what angle of incidence does an equilateral prism of index $1.5$ give minimum deviation?
  • 7 For the prism of part 6, find the smallest angle of incidence for which a ray emerges from the second face.
  • 8 A thin prism of angle $6.0^\circ$ ($n=1.50$) is combined with a thin prism of index $1.75$ so that the combination produces no deviation. Find the angle of the second prism.
1 $\sin r_1=\tfrac{\sin50^\circ}{1.5}=0.511$, $r_1\approx30.7^\circ$; $r_2\approx29.3^\circ$; $\sin e=1.5\sin29.3^\circ=0.734$, $e\approx47.2^\circ$; $\delta\approx50+47.2-60=37.2^\circ$ (very close to minimum deviation).
2 $n=\dfrac{\sin45^\circ}{\sin30^\circ}=\sqrt2\approx1.41$.
3 $\sin\dfrac{60^\circ+\delta_m}{2}=\sqrt3\times\tfrac12$, so $\dfrac{60^\circ+\delta_m}{2}=60^\circ$ and $\delta_m=60^\circ$.
4 $\delta=(1.6-1)\times5.0^\circ=3.0^\circ$.
5 $i=0$, $r_1=0$, $r_2=30^\circ$; $\sin e=1.5\times0.5=0.75$, $e\approx48.6^\circ$; $\delta=0+48.6-30=18.6^\circ$.
6 $\sin i=1.5\sin30^\circ=0.75$, $i\approx48.6^\circ$.
7 Emergence needs $r_2\le C=41.8^\circ$, so $r_1\ge18.2^\circ$: $\sin i\ge1.5\sin18.2^\circ=0.468$, $i\ge27.9^\circ$.
8 Opposed prisms: $(n_1-1)A_1=(n_2-1)A_2$, so $A_2=\dfrac{0.50\times6.0^\circ}{0.75}=4.0^\circ$.
II
The eye and the simple microscope. $D=25$ cm.
  • 1 Treat the eye as a single lens with the retina $2.0$ cm behind it. Find its power when viewing (a) a distant object and (b) an object at the near point. What is the range of accommodation?
  • 2 Find the magnifying power of a magnifier of focal length $10$ cm with the image (a) at the near point and (b) at infinity.
  • 3 A magnifier gives a magnifying power of $6.0$ with the image at the near point. Find its focal length.
  • 4 Where should an object be placed in front of a magnifier of focal length $5.0$ cm for the image to be at the near point?
  • 5 Why can a single lens not usefully magnify more than about ten times?
  • 6 An elderly person's near point is $50$ cm. Using a magnifier of focal length $10$ cm with the image at that near point, what magnifying power do they get, compared with their own best unaided view?
  • 7 An insect $1.0$ mm long is viewed (a) with the unaided eye at the near point and (b) through a magnifier of focal length $5.0$ cm with the image at infinity. Find the angle subtended in each case.
1 $v=+0.020$ m. (a) $u\to-\infty$: $P=\tfrac1v=50$ D. (b) $u=-0.25$ m: $P=\tfrac1{0.020}+\tfrac1{0.25}=54$ D. Range of accommodation $4.0$ D.
2 (a) $1+\tfrac{25}{10}=3.5$; (b) $\tfrac{25}{10}=2.5$.
3 $1+\tfrac{25}{f}=6.0$, $f=5.0$ cm.
4 $\tfrac1u=-\tfrac1{25}-\tfrac15=-\tfrac6{25}$, $u\approx-4.2$ cm: $4.2$ cm from the lens.
5 A higher power needs a shorter focal length, i.e. a very strongly curved, tiny lens; such lenses distort and colour-fringe the image badly, and the object must be held impractically close.
6 With $D'=50$ cm: $m=1+\tfrac{50}{10}=6$ — their unaided view is at $50$ cm, so the gain is larger than for a normal eye.
7 (a) $\tfrac{1.0}{250}=4.0\times10^{-3}$ rad. (b) $\tfrac{1.0}{50}=2.0\times10^{-2}$ rad — five times larger.
III
The compound microscope. $D=25$ cm.
  • 1 A microscope has $f_o=0.50$ cm, $f_e=2.0$ cm and a tube length of $16$ cm. Estimate its magnifying power in normal adjustment.
  • 2 An object is $0.60$ cm from an objective of focal length $0.50$ cm; the eyepiece has $f_e=2.5$ cm and the final image is at the near point. Find the magnifying power.
  • 3 For the microscope of part 2, find the separation of the lenses if the final image is instead at infinity.
  • 4 The objective of a microscope gives $m_o=-30$. What eyepiece focal length gives a magnifying power of $300$ in normal adjustment?
  • 5 An objective forms an image with $m_o=-40$; the eyepiece has $f_e=5.0$ cm and the final image is at the near point. Find the magnifying power.
  • 6 Explain why the final image in a compound microscope is inverted.
  • 7 Why must both lenses of a compound microscope have short focal lengths?
1 $|m|\approx\dfrac{16}{0.50}\times\dfrac{25}{2.0}=32\times12.5=400$.
2 $\tfrac1{v_o}=\tfrac1{0.50}-\tfrac1{0.60}=\tfrac13$, $v_o=3.0$ cm, $m_o=-5.0$; $m_e=1+\tfrac{25}{2.5}=11$; $m=-55$ (magnifying power $55$, inverted).
3 Intermediate image at the eyepiece's focus: $3.0+2.5=5.5$ cm.
4 Need $m_e=10=\tfrac{25}{f_e}$: $f_e=2.5$ cm.
5 $m_e=1+\tfrac{25}{5.0}=6$; $m=-40\times6=-240$.
6 The objective forms a real, inverted image; the eyepiece, acting as a magnifier, keeps that image's orientation, so the final image is inverted relative to the object.
7 $m_o\approx-L/f_o$ grows as $f_o$ shrinks, and $m_e=D/f_e$ grows as $f_e$ shrinks; the overall power is their product.
IV
Telescopes and applications.
  • 1 An astronomical telescope has $f_o=150$ cm and $f_e=5.0$ cm. Find its magnifying power and length in normal adjustment.
  • 2 Find the magnifying power of the telescope of part 1 when the final image is at the near point.
  • 3 The Moon subtends $0.50^\circ$. What angle does it subtend through the telescope of part 1 in normal adjustment?
  • 4 Find the diameter of the Moon's image formed by the objective of part 1.
  • 5 Why do astronomical telescopes have objectives of long focal length and large diameter?
  • 6 Give three advantages of a reflecting telescope over a refracting telescope of the same aperture.
  • 7 Modelling. A birdwatcher wants a telescope no longer than $40$ cm with a magnifying power of at least $15$ in normal adjustment. Find the largest eyepiece focal length she can use, and the matching objective focal length.
1 $m=\tfrac{150}{5.0}=30$; length $155$ cm.
2 $m=30\left(1+\tfrac{5.0}{25}\right)=36$.
3 $30\times0.50^\circ=15^\circ$.
4 $h=f_o\alpha=150\times(0.50\times\tfrac{\pi}{180})\approx1.3$ cm.
5 Long $f_o$ gives large $m=f_o/f_e$; a large diameter collects more light, so faint objects become visible (and finer detail can be seen).
6 No chromatic aberration (reflection does not depend on colour); a paraboloidal mirror has no spherical aberration; a mirror can be supported over its whole back and made in segments, so it can be much larger and lighter; only one surface needs to be figured precisely.
7 $f_o=15f_e$ and $f_o+f_e=16f_e\le40$ cm: $f_e\le2.5$ cm, with $f_o=37.5$ cm. (Real birdwatching telescopes add prisms to make the image erect.)
🎯 Exam corner — prisms and optical instruments BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks Draw a ray diagram for the refraction of a ray through a glass prism and show the angle of deviation. Show that the refractive index of the prism is $n=\dfrac{\sin\frac{A+\delta_m}{2}}{\sin\frac A2}$, where $\delta_m$ is the angle of minimum deviation.

Geometry In the quadrilateral formed by the edge, the two points of incidence and the meeting point $N$ of the normals, two angles are $90^\circ$, so the angle at $N$ is $180^\circ-A$. In the triangle formed by the ray inside and the two normals: $r_1+r_2=A$.
Deviation $\delta=(i-r_1)+(e-r_2)=i+e-A$.
Minimum At minimum deviation the path is symmetric: $i=e$, $r_1=r_2=A/2$, and $\delta_m=2i-A$, so $i=\dfrac{A+\delta_m}{2}$. Snell's law at the first face: $n=\dfrac{\sin i}{\sin r}=\dfrac{\sin\frac{A+\delta_m}{2}}{\sin\frac A2}$.
Exam tip Mark $i$, $e$, $r_1$, $r_2$, $A$ and $\delta$ on the diagram; the reason for $i=e$ (reversibility of light, unique minimum) earns credit.

E2. Board · 1 mark Assertion (A): The world's largest optical telescopes use concave mirrors rather than lenses as objectives. Reason (R): A mirror has no chromatic aberration and can be supported over its whole back surface. (A) Both A and R are true and R is the correct explanation of A. (B) Both are true but R is not the correct explanation of A. (C) A is true, R is false. (D) A is false, R is true.

Concept Large lenses sag under their own weight (supported only at the edge) and focus colours differently; mirrors avoid both.
Answer (A).

E3. JEE Main For an equilateral prism, the angle of minimum deviation is found to be equal to the angle of the prism. The refractive index of the prism is (A) $1.41$ (B) $1.50$ (C) $1.73$ (D) $2.00$.

Steps $\delta_m=A=60^\circ$: $n=\dfrac{\sin60^\circ}{\sin30^\circ}=\sqrt3\approx1.73$.
Answer (C). Exam tip: here $i=\dfrac{A+\delta_m}{2}=60^\circ$ — the ray enters at the prism angle itself.

E4. NEET A compound microscope has an objective of focal length $1.0$ cm, an eyepiece of focal length $5.0$ cm and a tube length of $20$ cm. Its magnifying power in normal adjustment ($D=25$ cm) is about (A) $50$ (B) $100$ (C) $120$ (D) $200$.

Formula $|m|\approx\dfrac{L}{f_o}\cdot\dfrac{D}{f_e}=\dfrac{20}{1.0}\times\dfrac{25}{5.0}=100$.
Answer (B). Exam tip: (C) is $20\times(1+5)$, the near-point value — read which adjustment is asked.

E5. JEE Advanced (One or more correct.) A prism of angle $60^\circ$ is made of glass of refractive index $\sqrt2$. (A) The angle of minimum deviation is $30^\circ$. (B) At minimum deviation the angle of incidence is $45^\circ$. (C) A ray emerges from the second face for every angle of incidence between $0^\circ$ and $90^\circ$. (D) The deviation for grazing incidence ($i=90^\circ$) is about $51.5^\circ$, the same as for grazing emergence.

(A), (B) $\sin\dfrac{60^\circ+\delta_m}{2}=\sqrt2\times\tfrac12$, so $\dfrac{60^\circ+\delta_m}2=45^\circ$: $\delta_m=30^\circ$ and $i=45^\circ$. Both true.
(C) $C=45^\circ$, so emergence needs $r_2\le45^\circ$, $r_1\ge15^\circ$, $\sin i\ge\sqrt2\sin15^\circ=0.366$, $i\ge21.5^\circ$. For smaller $i$ the ray is totally reflected: false.
(D) $i=90^\circ$: $r_1=45^\circ$, $r_2=15^\circ$, $\sin e=0.366$, $e\approx21.5^\circ$, $\delta=90+21.5-60\approx51.5^\circ$. By reversibility grazing emergence ($e=90^\circ$, $i=21.5^\circ$) gives the same: true.
Answer (A), (B), (D). Exam tip: the $\delta$–$i$ curve starts and ends at the same height — check it in Interactive 1.

E6. Board · Case study A school astronomy club builds a refracting telescope with an objective of focal length $1.2$ m and an eyepiece of focal length $4.0$ cm. (i) (2 marks) Find its magnifying power in normal adjustment and the distance between the lenses. (ii) (1 mark) Jupiter subtends about $45$ seconds of arc ($2.2\times10^{-4}$ rad). What angle does it subtend through the telescope? (iii) (1 mark) State one reason why the club might prefer a reflecting telescope of the same aperture.

(i) $m=\dfrac{120}{4.0}=30$; separation $=124$ cm.
(ii) $\beta=m\alpha=30\times2.2\times10^{-4}=6.6\times10^{-3}$ rad ($\approx0.38^\circ$, about $23$ minutes of arc — comparable to the Moon seen with the unaided eye).
(iii) No chromatic aberration (no coloured fringes round Jupiter), and a mirror is cheaper and lighter than a lens of the same diameter. Exam tip: one clearly stated reason is enough for one mark.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Can a hollow glass prism filled with fruit juice measure its sugar content through the angle of minimum deviation — and how precisely must the angle be read?

Why it matters. Sugar mills in India and Brazil, fruit growers, honey producers and wine makers check sugar content with refractometers, which measure refractive index. Dissolved sugar raises the refractive index of water, so a measurement of $n$ reads out the sugar concentration. A minimum-deviation prism is the simplest refractometer you can build.

What to do.

  1. Use these illustrative values for sugar solutions: $n=1.333$ (pure water), $1.348$ ($10\%$ sugar by mass), $1.364$ ($20\%$). For a hollow prism of angle $60^\circ$ (thin walls do not deviate the light), find $\delta_m$ for each, using $\delta_m=2\sin^{-1}\left(n\sin\tfrac A2\right)-A$.
  2. Estimate how much $\delta_m$ changes per $1\%$ of sugar.
  3. A school spectrometer reads angles to $0.1^\circ$ (6 minutes of arc). What is the smallest change in sugar content it can detect? Would a professional refractometer that measures $n$ to $\pm0.0002$ do better?
  4. Would a prism of angle $30^\circ$ be more or less sensitive? Calculate and explain.

Tools: Calculator or spreadsheet; Interactive 1 ("hollow prism of water") to see the deviation curve.

Hand in: A table of $\delta_m$ against concentration, the sensitivity in degrees per percent, the detection limit and a paragraph comparing the two prism angles.

Going further. Find out how a hand-held refractometer uses the critical angle (Section 9.2) instead of minimum deviation, and why that makes it work with cloudy juices.

Instructor note: $A=60^\circ$: $\delta_m\approx23.6^\circ,\ 24.7^\circ,\ 26.0^\circ$ for $0,\ 10,\ 20\%$; about $0.12^\circ$ per $1\%$, so a $0.1^\circ$ reading detects roughly $1\%$. $\pm0.0002$ in $n$ corresponds to about $\pm0.13\%$ sugar (since $n$ rises by about $0.0015$ per $1\%$) — almost ten times better. $A=30^\circ$: $\delta_m\approx10.4^\circ,\ 10.8^\circ,\ 11.4^\circ$, about $0.05^\circ$ per $1\%$ — less sensitive, because a smaller prism angle gives a smaller deviation for the same index change. The refractive-index values are illustrative.

🧪 Self-check quiz

0/5 correct

Q1. When a ray passes through a prism at minimum deviation,

Q2. A thin prism of angle $6^\circ$ and refractive index $1.5$ deviates a ray through

Q3. A magnifying glass of focal length $5$ cm is used with the image at the near point ($25$ cm). Its magnifying power is

Q4. An astronomical telescope in normal adjustment has $f_o=100$ cm and $f_e=5$ cm. Its magnifying power and length are

Q5. To increase the magnifying power of a compound microscope you should

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