"Optics" — Practice

Question 1

A glass sphere of radius $R$ and refractive index $1.5$ has a parallel beam of light incident on it. Find where the beam is focused, measured from the far surface.

Accepted Answer

First surface, $u=\infty$: $\dfrac{1.5}{v}=\dfrac{0.5}{R}\qquad\Rightarrow\qquad v=3R$ from the first surface.

This is $3R-2R=R$ beyond the second surface, so it is a **virtual object** at $u=+R$.

Second surface: $\dfrac{1}{v}-\dfrac{1.5}{R}=\dfrac{1-1.5}{-R}=\dfrac{0.5}{R}$

$\dfrac1v=\dfrac{2}{R}\qquad\Rightarrow\qquad v=\dfrac R2$ beyond the sphere.

Question 2

A thin equiconvex lens of index $1.5$ has focal length $20$ cm in air. Find its focal length when immersed in a liquid of index $1.6$, and describe the change.

Accepted Answer

In air: $\dfrac{1}{20}=\left(0.5ight)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}ight)$, so the bracket equals $\dfrac{1}{10}$ per cm.

In the liquid: $\dfrac{1}{f}=\left(\dfrac{1.5}{1.6}-1ight)\left(\dfrac{1}{10}ight)=\left(-0.0625ight)\left(0.1ight)$

$f=-160$ cm

The lens becomes **diverging**, because it is now optically rarer than its surroundings.

Question 3

An object and a screen are $90$ cm apart. A lens placed between them gives sharp images at two positions $30$ cm apart. Find the focal length and both magnifications.

Accepted Answer

$f=\dfrac{D^{2}-d^{2}}{4D}=\dfrac{8100-900}{360}=20$ cm

$m_1=\dfrac{D-d}{D+d}=\dfrac{60}{120}=0.5$ and $m_2=\dfrac{120}{60}=2$

Their product is $1$, as it must be, since the two positions simply exchange the object and image distances.

Question 4

A thin plano-convex lens of index $1.5$ and radius of curvature $30$ cm has its plane face silvered. Find the focal length of the equivalent mirror.

Accepted Answer

$\dfrac{1}{f_{lens}}=\left(0.5ight)\left(\dfrac{1}{30}ight)=\dfrac{1}{60}$, so $f_{lens}=60$ cm.

The silvered plane face has $R=\infty$, so $P_{mirror}=0$.

$P=2\left(\dfrac{1}{60}ight)+0=\dfrac{1}{30}$, giving $f=30$ cm, converging.

Silvering the plane face therefore halves the effective focal length of the lens.

Question 5

A prism of angle $30^{\circ}$ and index $1.5$ has light incident at $45^{\circ}$ on one face. Find the angle of emergence and the deviation.

Accepted Answer

$\sin r_1=\dfrac{\sin45^{\circ}}{1.5}=0.471\qquad\Rightarrow\qquad r_1=28.1^{\circ}$

$r_2=A-r_1=30-28.1=1.9^{\circ}$

$\sin e=1.5\sin1.9^{\circ}=1.5	imes0.0332=0.0498\qquad\Rightarrow\qquad e=2.9^{\circ}$

$\delta=i+e-A=45+2.9-30=17.9^{\circ}$

Question 6

Two thin prisms are to be combined to give dispersion without deviation. The first has angle $6^{\circ}$ with $n_y=1.52$ and $\omega=0.021$; the second has $n'_y=1.65$. Find the required angle of the second prism.

Accepted Answer

For no net deviation of the mean ray:

$\left(n_y-1ight)A+\left(n'_y-1ight)A'=0$

$\left(0.52ight)(6)+\left(0.65ight)A'=0$

$A'=-\dfrac{3.12}{0.65}=-4.8^{\circ}$

The negative sign means the second prism must be inverted relative to the first, which is exactly how a direct-vision spectroscope is built.

Question 7

An object moves along the axis towards a convex lens of focal length $15$ cm at $3$ cm s$^{-1}$, and is currently $45$ cm from it. Find the speed and direction of the image.

Accepted Answer

Using $\dfrac1v-\dfrac1u=\dfrac1f$ with $u=-45$ cm and $f=+15$ cm:

$\dfrac1v=\dfrac{1}{15}-\dfrac{1}{45}=\dfrac{2}{45}\qquad\Rightarrow\qquad v=22.5$ cm

$\left|might|=\dfrac{v}{\left|uight|}=\dfrac{22.5}{45}=0.5$

$v_{image}=m^{2}v_{object}=0.25	imes3=0.75$ cm s$^{-1}$

The image moves **away** from the lens as the object approaches it.

Question 8

In a double-slit experiment with $\lambda=500$ nm, $d=1$ mm and $D=1$ m, the whole apparatus is immersed in water of index $1.33$. Find the new fringe width and explain the change.

Accepted Answer

In air: $\beta=\dfrac{\left(5	imes10^{-7}ight)(1)}{10^{-3}}=5	imes10^{-4}$ m $=0.5$ mm

In water the wavelength becomes $\dfrac{500}{1.33}=376$ nm while $d$ and $D$ are unchanged:

$\beta'=\dfrac{0.5}{1.33}=0.376$ mm

The fringes crowd together because the wavelength inside the medium is shorter; the geometry is untouched.

Question 9

A thin sheet of index $1.6$ and thickness $2\ \mu$m covers one slit of a double-slit apparatus with $\lambda=600$ nm. How many fringes does the pattern shift, and in which direction?

Accepted Answer

Extra optical path $=\left(n-1ight)t=\left(0.6ight)\left(2	imes10^{-6}ight)=1.2	imes10^{-6}$ m

Number of fringes $=\dfrac{1.2	imes10^{-6}}{6	imes10^{-7}}=2$

The pattern shifts by exactly **two fringe widths, towards the covered slit**, because that path now needs less geometric distance to reach equality.

The fringe width itself is unchanged.

Question 10

An oil film of index $1.45$ floats on water of index $1.33$ and is viewed normally in white light. Find the smallest thickness for which $580$ nm is strongly reflected.

Accepted Answer

The top reflection (air to oil) is off a denser medium and gains $\pi$; the bottom reflection (oil to water) is also off a denser medium and gains $\pi$ as well.

The two phase changes **cancel**, so the ordinary path condition applies:

$2\mu t=n\lambda$, with $n=1$:

$t=\dfrac{580}{2	imes1.45}=200$ nm

Had the film been on glass of higher index than oil, the answer would have been the half-wavelength condition instead.

Question 11

Light of $650$ nm falls on a single slit of width $0.15$ mm, with a screen $1.8$ m away. Find the width of the central maximum and the position of the second minimum.

Accepted Answer

Central maximum width $=\dfrac{2\lambda D}{a}=\dfrac{2\left(6.5	imes10^{-7}ight)\left(1.8ight)}{1.5	imes10^{-4}}$

$=1.56	imes10^{-2}$ m $=15.6$ mm

Second minimum: $y_2=\dfrac{2\lambda D}{a}=15.6$ mm from the centre.

The first minimum is at $7.8$ mm, so the central maximum spans from $-7.8$ mm to $+7.8$ mm — twice the width of each side maximum.

Question 12

The human eye has a pupil about $2$ mm across. Estimate the smallest separation it can resolve at $25$ cm for $550$ nm light, and compare with the typical spacing between retinal cells.

Accepted Answer

$	heta_{min}=\dfrac{1.22	imes5.5	imes10^{-7}}{2	imes10^{-3}}=3.36	imes10^{-4}$ rad

Separation $=	heta_{min}	imes0.25=8.4	imes10^{-5}$ m, about $0.08$ mm.

That is roughly the spacing of cone cells on the retina, which is not a coincidence: there would be no advantage in packing detectors more finely than diffraction allows.

Practice — "Optics"

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