Ray Optics and Optical Test

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Ray Optics and Optical Test - 11

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Question 1
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The magnifying power of a telescope with tube length $$60cm$$ is $$5$$. What is the focal length of its eye piece?

A
$$30cm$$

B
$$20cm$$

C
$$10cm$$

D
$$40cm$$

Solution

The magnifying power of a telescope is given by

$$m = \dfrac{f_0}{f_e}, \, 5 = \dfrac{f_0}{f_e}$$

$$f_0 + f_e = 60$$ [The total length of the telescope is $$60cm$$]

And $$f_0 = 5f_e$$

$$\therefore f_0 + f_e = 5f_0, \, 6f_e = 60, f_e = 10cm$$

$$f_0 = 5f_e = 50cm$$

$$\therefore$$ The focal length of the eyepiece is $$f_e = 10cm$$
Question 2
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The refractive index of the material of a concave lens is $$\mu.$$ It is immersed in a medium of refractive index $$\mu_1$$. A parallel beam of light is incident on the lens. The path of the emergent rays when $$\mu_1>\mu$$ is :

A

B

C

D

Solution

From the laws of optics, we know that a lens reverses its behaviour when placed in a medium of higher refractive index. Hence, a concave lens diverges light in general, but when placed in a medium of higher refractive index it converges parallel rays of light to a focal point.
Question 3
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A ray of light is incident on a prism $$\mathrm{A}BC$$ of refractive index $$\sqrt{3}$$ as shown in figure.
(a) Find the angle of incidence for which the deviation of light ray by the prism $$\mathrm{A}BC$$ is minimum.
(b) By what angle the second prism must be rotated, so that the final ray suffer net minimum deviation.

A
$$30^o$$, $$60^o$$

B
$$60^o$$, $$30^o$$

C
$$60^o$$, $$60^o$$

D
None of these

Solution

Minimum deviation
$$i_{1}=i_{2}=i$$ and $$r_{1}=r_{2}=r$$

and $$r_{1}+r_{2}=A$$

$$r=A/2=30^{\circ}$$

Using Snell's law,

$$\dfrac{sini_{1}}{sinr_{1}}=\mu$$

$$sini_{1}=\sqrt{3}sin30^{\circ}$$

$$sini_{1}=\dfrac{\sqrt{3}}{2}$$

$$i_{1}=60^{\circ}$$

When the prism is rotated anticlockwise by $$60^{\circ}$$ the combined prism will become a slab as shown in the figure 2. In this case the net deviation by the prism will be zero as the deviation by the two prism will be equal and opposite.
Question 4
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The aperture diameter of a telescope is $$5m$$. The separation between the moon and the earth is $$4\times 10^5$$ km. With light of wavelength of $$5500\overset{o}{A}$$, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to?

A
$$60$$ m

B
$$20$$ m

C
$$600$$ m

D
$$200$$ m

Solution

Refer above image.

Let the aperture $$=d$$ $$D=4\times 10^8m$$

$$wavelength =\lambda=5500\times 10^{-10}m$$

for two objects to be resolved

$$\theta=1.22\dfrac{\lambda}{d}$$

also $$d=D\theta$$

$$\Rightarrow \theta=\dfrac{d}{D}$$

$$\Rightarrow \dfrac{d}{4\times 10^8}=\dfrac{1.22\times 5500\times 10^{-10}}{5}$$

$$d=53.68\approx 60m$$

Option A is correct.
Question 5
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There is a small source of light at some depth below the surface of water (refractive index $$=4/3$$) in a tank of large cross sectional surface area. Neglecting any reflection from the bottom and the absorption by water, percentage of light that emerges out of surface is (nearly):
[Use the fact that surface area of a spherical cap of height $$h$$ and radius of curvature $$r$$ is $$2\pi r h$$]

A
$$21\%$$

B
$$34\%$$

C
$$17\%$$

D
$$50\%$$

Solution

$$\sin\theta=\dfrac{1}{\mu}$$ here angle is $$\beta$$
$$\Rightarrow \dfrac{1}{\mu}=\sin \beta=\dfrac{1}{4/3}$$

$$\Rightarrow \sin \beta=\dfrac{3}{4}$$
$$\Rightarrow \cos \beta=\dfrac{\sqrt{7}}{4}$$
We know
Solid angle, $$d\Omega=2\pi R^2(1-\cos\beta)$$
Percentage of light $$=\dfrac{2\pi R^2(1-\cos\beta)}{4\pi R^2}\times 100$$
$$=\dfrac{1-\cos\beta}{2}\times 100=\left(\dfrac{4-\sqrt{7}}{8}\right)\times 100$$
$$\boxed{=17\%}$$
Question 6
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Monochromatic light is incident on a glass prism of angle A. If the refractive index of the material of the prism is $$\mu$$, a ray incident at an angle $$\theta$$, on the face AB would get transmitted through the face AC of the prism provided

A
$$\theta > sin^{-1}\left[\mu \, sin \left(A - sin^{-1}\left(\frac{1}{\mu}\right) \right)\right]$$

B
$$\theta < sin^{-1}\left[\mu \, sin \left(A - sin^{-1}\left(\frac{1}{\mu}\right) \right)\right]$$

C
$$\theta > cos^{-1}\left[\mu \, sin \left(A + sin^{-1}\left(\frac{1}{\mu}\right) \right)\right]$$

D
$$\theta < cos^{-1}\left[\mu \, sin \left(A + sin^{-1}\left(\frac{1}{\mu}\right) \right)\right]$$

Solution

Consider refraction though AB surface

Using Snell's law,

$$1\sin\theta=\mu \sin r$$...(i)

Now for ray to transmit through AC the angle of incidence should be less the

critical angle

i.e. $$ r'<\sin^{-1}\dfrac{1}{\mu}$$

From triangle APQ$$ A+90^\circ{}-r+90^\circ{}-r'=180$$

$$r'=A-r;$$ $$A-r<\sin^{-1}\dfrac{1}{\mu}$$

From equation (i)

$$\theta > sin^{-1}\left[\mu \, sin \left(A - sin^{-1}\left(\dfrac{1}{\mu}\right) \right)\right]$$

Option A is correct.
Question 7
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A parallel beam of light is incident from air at an angle $$\alpha $$ on the side PQ of a right-angled triangular prism of refractive index $$n=\sqrt { 2 } $$. Light undergoes total internal reflection in the prism at the face PR when $$\alpha $$ has minimum value of $${ 45 }^{ \circ }$$. The angle $$\theta $$ of the prism is:

A
$${ 15 }^{ \circ }$$

B
$${ 22.5 }^{ \circ }$$

C
$${ 30 }^{ \circ }$$

D
$${ 45 }^{ \circ }$$

Solution

Applying Snell's law at surface PQ,
$$\therefore 1\times \sin { \alpha } =n\times \sin { \beta } $$
at $$\alpha =45°,n=\sqrt { 2 } $$
$$\sin { \beta } =\cfrac { 1 }{ 2 } \therefore \beta =30°$$
$$\gamma =90-\left( 180-\left( 90+\theta +\beta \right) \right) $$
$$\therefore $$ Applying Snell's law at surface PR
$$\therefore \sqrt { 2 } \sin { \gamma } =1$$
$$\therefore \sin { \left( \theta +\beta \right) } =\cfrac { 1 }{ \sqrt { 2 } } $$
$$\therefore \theta +\beta =45°$$
$$\therefore \theta +30°=45°$$
$$\therefore \theta =15°$$ Thus option A
Question 8
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A beam of light consisting of red green and blue colours is incident on a right angled prism. The refractive index of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47, respectively. The prism will:

A
separate the red colour part from the green and blue colours

B
separate the blue colour part front the red and green colours

C
separateall all three colours from one another

D
not separate the three colours at all

Solution

For total internal reflection, $$i \gt i_c$$
Here $$i=45^0$$
$$\frac{\sin i}{\sin r} =\mu$$
$$\sin 45^0 \gt \frac{1}{\mu} $$
$$ \mu \gt \frac{1}{\sqrt{2}}=1.414 $$
$$\mu_{red} \lt 1.414$$ but $$\mu _{green} \gt 1.414$$ and $$\mu _{violet} \gt 1.414$$.
Hence, green and violet will be totally internally reflected.
Red will be refracted.
Question 9
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An astronomical refracting telescope will have large angular magnification and high resolution, when it has an objective lens of :

A
Large focal length and large diameter

B
Small focal length and large diameter

C
Small focal length and small diameter

D
Large focal length and small diameter

Solution
Question 10
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A convex lens of refractive index $$\dfrac{3}{2}$$ has a power of 2.5 D in air. If it is placed in a liquid of refractive index 2, then the new power of the lens is:

A
- 1.25 D

B
- 1.5 D

C
1.25 D

D
1.5 D

Solution

According to Lens Maker's Formula,
Power of lens$$=P=(\mu_r-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})$$
where $$\mu_r$$ is the refractive index of material of lens relative to the surrounding medium$$=\dfrac{\mu_{material}}{\mu_{medium}}$$
For convex lens in air, $$\mu_r=\dfrac{3/2}{1}=\dfrac{3}{2}$$
For convex lens in the liquid, $$\mu_r=\dfrac{3/2}{2}=\dfrac{3}{4}$$
Thus $$\dfrac{P_2}{P_1}=\dfrac{\mu_{r_2}-1}{\mu_{r_1}-1}$$
$$=\dfrac{\dfrac{3}{4}-1}{\dfrac{3}{2}-1}=-\dfrac{1}{2}$$
$$\implies P_2=-\dfrac{1}{2}\times 2.5D=-1.25D$$