Ray Optics and Optical Test

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Ray Optics and Optical Test - 62

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Question 1
1 / -0

In the figure ABC is the cross section of a right angled prism and BCDE is the cross section of a glass slab. The value of $$\theta$$ so that light incident normally on the face AB does not cross the face BC is (given sin$$^{-1}$$ (3/5) $$=$$ 37$$^o$$)

A
$$\theta \leq 37^o$$

B
$$\theta > 37^o$$

C
$$\theta \leq 53^o$$

D
$$\theta < 53^o$$

Solution

For no refraction at surface BC

$$\dfrac { sin(i) }{ sin(r) } =\dfrac { { \mu }_{ 2 } }{ { \mu }_{ 1 } } \\ for\quad no\quad refraction\quad at\quad surface\quad BC\quad$$

$$r\ge { 90 }^{ 0 }$$
here,
$${ \mu }_{ 1 }=\dfrac { 3 }{ 2 } \ (refractive\ index\ of\ ABC)\\$$

$$ { \mu }_{ 2 }=\dfrac { 6 }{ 5 } \ \ \ (refractive\ index\ of\ BCDE)$$
thus
$$\dfrac { sin(i) }{ sin(90) } =\dfrac { \dfrac { 6 }{ 5 } }{ \dfrac { 3 }{ 2 } } =\dfrac { 4 }{ 5 } \\$$
$$ \Rightarrow sini\ge \dfrac { 4 }{ 5 } \\$$
$$ \Rightarrow i\ge { 53 }^{ 0 } $$
here in triangle ABC

$$i={ 90 }^{ 0 }-\theta \\ \Rightarrow { 90 }^{ 0 }-\theta \ge { 53 }^{ 0 }\\ \theta \le { 3 }7^{ 0 }$$
Question 2
1 / -0

Rising and setting sun appears to be reddish because:

A
The sun is colder at sunrise or at sunset

B
Diffraction sends red rays to the earth at these times

C
Refraction is responsible for it

D
Scattering due to dust particles and air molecules is responsible for it

Solution

According to Rayleigh's law of scattering, amount of scattering $$\alpha \frac {1}{\lambda^4}$$.
Red having maximum wavelength straight to the observer.
Most of the scattered light goes towards the sky, so the sky appears blue.
Question 3
1 / -0

A ray incident at a point at an angle of incidence $$\theta$$ enters into a glass sphere placed in air which is reflected and refracted at the farther surface of the sphere as shown in the figure. The angle between reflected and refracted rays at this surface is $$90^{0}$$. If refractive index of the sphere is $$\sqrt{3} ,$$ the angle $$\theta$$ is :

A
$$\dfrac{\pi }{3}$$

B
$$\dfrac{\pi }{4}$$

C
$$\dfrac{\pi }{6}$$

D
$$\dfrac{2\pi }{3}$$

Solution

Applying Snell's law on the 2 surfaces,

$$ sin \theta = \sqrt{3} sin r $$

Where, r is the angle of refraction.

Applying on the second surface,
$$ sin r = cos \theta $$

Equating the 2 terms,

$$ sin \theta = \sqrt{3} cos \theta $$

Thus, $$ tan \theta = \sqrt{3} $$

Or, $$ \theta = 60^0 $$
Question 4
1 / -0

A glass prism of $$\mu=$$ 1.5 is immersed in water of $$ [\mu= \frac{4}{3}]$$ as shown in this figure. A ray of light incident normally on face AB is totally reflected at the face AC if $$sin\theta$$ is

A
less than or equal to $$\frac{2}{3}$$

B
greater than $$ \frac{8}{9}$$

C
$$0.8666$$

D
between $$ \frac{2}{4} \ and \ \frac{8}{9}$$

Solution

Total internal reflection should be at point P, so that all the rays can be reflected.
For total internal reflection incident agnle on face AC should be grater than the crictical angle.
For crictical angle -
Apply snell's law:
$$ \mu_{1} sin\theta \ = \mu_2 sin90$$
$$ \implies \ 1.5 sin\theta = \frac{4 }{3} $$
$$ \implies sin\theta = \frac{8 }{9} $$
For total internal reflection -
$$ sin\theta > \frac{8}{9}$$
Hence, optionB is correct answer.
Question 5
1 / -0

The angle of crown glass $$(\mu=1.52)$$ prism is $$5^o$$. What should be the angle of flint glass $$(\mu=1.63)$$ prism so that the two prisms together may be used in a direct vision spectroscope?

A
$$-2.14^o$$

B
$$+2.14^o$$

C
$$-4.12^o$$

D
$$+4.12^o$$

Solution

For direct vision spectroscope,
$$\delta_1+\delta_2=0$$

$$\therefore (\mu_1-1)A_1+(\mu_2-1)A_2=0$$

$$\Rightarrow A_2=-[\dfrac {\mu_1-1}{\mu_2-1}]A_1$$ $$=-[\dfrac {1.52-1}{1.63-1}]\times 5^0=-4.12^o$$
Question 6
1 / -0

The lens of a simple magnifier has a focal length of 2.5 cm. Calculate the angular magnification produced when the image is at the least distance of distinct vision.

A
10

B
5

C
8

D
15

Solution

Focal length of a magnifier
$$f=2.5 cm$$

(a) At least distance of distant vision, Angular magnification
$$=1+\dfrac {D}{f}=1+\dfrac {25 cm}{2.5 cm}$$

$$=11$$

(b) At infinity, angular magnification

$$=\dfrac {D}{f}=\dfrac {25 cm}{2.5 cm}=10$$
Question 7
1 / -0

A glass prism of refractive index $$1.5$$ is immersed in water (refractive index $$4/3)$$. A light beam normally on the face $$AB$$ is totally reflected to reach on the face $$BC$$ if

A
$$\sin\theta \geq \dfrac {8}{9}$$

B
$$\dfrac {2}{3} < \sin\theta \geq \dfrac {8}{9}$$

C
$$\sin\theta \leq \dfrac {2}{3}$$

D
$$\sin\theta \leq \dfrac {8}{9}$$

Solution

A light beam incident normally on the face AB is totally reflected to reach on the face BC if $$\theta\geq\theta_c$$
The beam is grazed on face BC, $$e=90^o$$
Refractive index of water,$$n_1=\dfrac{4}{3}$$ and that of prism,$$n_2=1.5$$
using Snell's law of refraction, $$n_2sin \theta_c=n_1sine$$
$$sin\theta_c=\dfrac{8}{9}$$
For total internal reflection $$sin\theta\geq sin\theta_c$$
Hence $$sin\theta \geq \dfrac{8}{9}$$
Question 8
1 / -0

A parallel beam of light falls on a solid transparent sphere. Which of the following options is/are correct?

A
If the beam is thick, then whole beam can be focussed at A.

B
The whole beam can be focussed at A only if the beam is thin enough.

C
If the beam is thin, then the beam can't be focussed before A.

D
None of these

Solution

Sphere is considered as a solid transparent with two
Spherical refractive surface of radius of curvature $$r_1=r_2=r$$
The image is formed at $$2r$$ distance from the pole of the 1st surface and so no refraction takes place in the second surface.
effectively refraction takes place in single surface and image is formed inside the refractive medium.
$$\therefore \dfrac{n_1}{u} +\dfrac{n_2}{v} =\dfrac{n_2 -n_1}{r}$$

$$n_1 = \text{refractive index of air}$$
$$n_2 = \text{refractive index of medium}=\mu$$
$$u=∞$$
$$v=2r$$

$$\dfrac{1}{\infty}+\dfrac{\mu}{2r}=\dfrac{(\mu-1)}{r}$$

$$\mu =2 \mu -2$$
$$\mu=2$$
Hence b is the correct answer.
Question 9
1 / -0

In Fig., $$ABC$$ is the cross section of a right-angled prism and $$ACDE$$ is the cross section of a glass slab. The value of $$\theta$$ so that light incident normally on the face $$AB$$ does not cross the face $$AC$$ is (given $$\sin^{-1} (3/5) = 37^{\circ})$$.

A
$$\theta \leq 37^{\circ}$$

B
$$\theta < 37^{\circ}$$

C
$$\theta \leq 53^{\circ}$$

D
$$\theta < 53^{\circ}$$

Solution

$$A = 90^{\circ} - \theta \Rightarrow r_{2} = A = 90^{\circ} - \theta > \theta_{c}$$
$$\cos \theta > \sin \theta_{c} = \dfrac {6/5}{2/3} = \dfrac {4}{5} (\theta_{C}$$ is critical angle)
$$\theta < \cos^{1} \dfrac {4}{5} = 37^{\circ}$$.
Question 10
1 / -0

A ray of light enters a spherical drop of water of refractive index $$\mu$$ as shown in fig. An expression of the angle between incident ray and emergent ray (angle of deviation) as shown in fig. is :

A
$$0^{\circ}$$

B
$$\phi$$

C
$$\alpha -\phi$$

D
$$\pi - 4\alpha + 2\phi$$

Solution

$$\alpha = (\phi - \alpha) + \theta$$
$$\delta = \pi - 2\theta = \pi - 4\alpha + 2\phi$$.