Optics — Practice

Question 1

Give the mirror and lens formulas together with their magnification expressions, and note the differences.

Accepted Answer

Mirror: $1/v+1/u = 1/f$ with $m = -v/u$. Lens: $1/v-1/u = 1/f$ with $m = v/u$. The sign of the second term and the presence of the minus in the magnification are both reversed between them, which is why mixing the two is a routine source of lost marks.

Question 2

Why does a diamond sparkle far more than a glass imitation of the same shape?

Accepted Answer

Diamond's refractive index of 2.42 gives a critical angle of only about $24^\circ$, so light entering it is totally internally reflected many times before finally escaping. Glass, with an index near 1.5, has a critical angle of about $42^\circ$ and traps far less light.

Question 3

Why can two identical light bulbs never produce an interference pattern?

Accepted Answer

Because interference requires coherent sources with a constant phase difference. Independent sources change phase randomly billions of times a second, so bright and dark regions shift far faster than any detector or eye can follow, and the pattern averages to uniform illumination.

Question 4

An object is 15 cm in front of a concave mirror of focal length 10 cm. Find the image position, magnification and nature.

Accepted Answer

With $u = -15$ and $f = -10$, the mirror formula gives $v = -30$ cm and $m = -2$. The image is 30 cm in front of the mirror, real, inverted and twice the size. For a mirror a negative $v$ means in front and therefore real — the opposite of the lens case.

Question 5

A glass lens of index 1.5 has focal length 20 cm in air. Find its focal length in a liquid of index 1.6.

Accepted Answer

In air the curvature bracket equals $1/10$. In the liquid the relative index is 0.9375, so $(n_{rel}-1) = -0.0625$, giving $f = -160$ cm. The converging lens has become diverging and much weaker, because it is now optically rarer than its surroundings.

Question 6

Slits 0.5 mm apart are 1 m from a screen, lit by 600 nm light. Find the fringe width, then find it with the apparatus immersed in water of index 4/3.

Accepted Answer

$\beta = \lambda D/d = 1.2$ mm. In water the wavelength falls by the factor $n$ while $D$ and $d$ are geometrical and unchanged, so the fringe width becomes 0.9 mm.

Question 7

Two slits have intensities in the ratio 4:1. Find the ratio of maximum to minimum intensity.

Accepted Answer

Using the amplitude ratio, $I_{max}/I_{min} = ((2+1)/(2-1))^{2} = 9$. The minima are not fully dark, because unequal amplitudes cannot cancel completely — which is why real experiments use identical slits.

Question 8

A point source is 2 m below water of index 4/3. Find the radius of the circle on the surface through which light escapes.

Accepted Answer

The critical angle has $\sin C = 3/4$, so $	an C = 3/\sqrt{7}$ and the radius is $h	an C \approx 2.27$ m. Beyond that circle light strikes the surface past the critical angle and is totally internally reflected, which is why a swimmer looking up sees a bright circular window in an otherwise mirrored surface.

Question 9

Light of 500 nm falls on a 0.2 mm slit with a screen 2 m away. Find the width of the central maximum and compare it with the other maxima.

Accepted Answer

The first minimum is at $\sin	heta = \lambda/a = 2.5	imes10^{-3}$, giving 5 mm on the screen, so the central maximum is 10 mm wide. Successive minima are 5 mm apart, so every other maximum is 5 mm wide. The central maximum is exactly twice as wide, which distinguishes diffraction from the uniform fringes of interference.

Question 10

Unpolarised light of intensity $I_0$ passes through two crossed polaroids with a third inserted between them at $30^\circ$ to the first. Find the emergent intensity.

Accepted Answer

The first polaroid gives $I_0/2$. The inserted one at $30^\circ$ gives $\cos^{2}30^\circ$ of that, which is $3I_0/8$. The final polaroid is at $60^\circ$ to the inserted one, giving $\cos^{2}60^\circ$ of that, so $3I_0/32$ emerges. Inserting a polaroid into a fully blocked path lets light through, because each one rotates the transmitted plane as well as attenuating it.

Question 11

An astronomical telescope has an objective of focal length 100 cm and an eyepiece of focal length 5 cm. Find its magnifying power in normal adjustment and the length of the tube. What would happen to both if the two lenses were swapped?

Accepted Answer

In normal adjustment $M = f_0/f_e = 100/5 = 20$, and the tube length is $f_0+f_e = 105$ cm. Swapping the lenses makes $M = 5/100 = 0.05$, so the instrument would shrink the image twentyfold instead of enlarging it, while the tube length stays 105 cm. A telescope therefore needs a long-focus objective and a short-focus eyepiece, and looking through the wrong end is exactly this swap.

Question 12

An object and a screen are fixed 100 cm apart. A converging lens placed between them forms a sharp image at two distinct positions, 40 cm apart. Find the focal length of the lens, and state the condition under which two such positions exist at all.

Accepted Answer

By the displacement method, $f = (D^{2}-d^{2})/4D = (100^{2}-40^{2})/400 = (10000-1600)/400 = 21$ cm. Two positions exist because object and image distances are interchangeable in the lens equation, so any solution has a mirror partner. They are distinct only when $D > 4f$; here $4f = 84$ cm, comfortably under 100 cm. At exactly $D = 4f$ the two positions merge into one and $d = 0$.

Practice — Optics

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