Ray Optics and Optical Test

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Ray Optics and Optical Test - 22

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Weekly Quiz Competition

Question 1
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A glass prism of refractive index $$1.5$$ is placed in water of refractive index $$1.33$$. The minimum value of the angle of the prism so that it will not be possible to have any emergent ray is :

A
$$150^{\circ}$$

B
$$125^{\circ}$$

C
$$165^{\circ}$$

D
$$180^{\circ}$$

Solution

$$\mu_{prism} =$$ $$1.5$$
$$\mu_{water}$$ $$=1.33$$

$$\mu_{r} =\dfrac{1.5}{1.33}$$

$$\theta _{cric} = sin^{-1} \left ( \dfrac{1}{\mu_{r}} \right )=sin^{-1}\left ( \dfrac{1.33}{1.5} \right )=62.5^{\circ}$$

Maximum possible angle A is given by: $$A=2\theta _{cric}=125^{\circ}$$
Question 2
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A double convex lens, made of a material of refractive index $$\mu _{1}$$ , is placed inside two liquids of refractive indices $$\mu _{2}$$ and $$\mu _{3}$$ , where $$\mu _{2}>\mu _{1}>\mu _{3}$$. A wide , parallel beam of light is incident on the lens from the left .The lens will give rise to

A
a single convergent beam

B
two different convergent beams

C
two different divergent beams

D
a convergent and a divergent beam

Solution

$$\therefore$$ analyzing the rays it gives a divergent ray above the axis and
convergent ray below $$\mu_2 > \mu_1 > \mu_3$$

from medium 2
$$\Rightarrow$$ when light ray enters into medium $$1 (\mu_1)$$ then moves away from normal

from medium 3
$$\Rightarrow$$ when light ray enters into medium $$1 (\mu_1)$$ the moves towards the normal.
Question 3
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Two equi-convex lenses each of focal lengths 20 cm and refractive index 1.5 are placed in contact and space between them is filled with water of refractive index 4/3. The combination works as :

A
converging lens of focal length 50 cm.

B
diverging lens of focal length 15 cm

C
converging lens of focal length 15 cm

D
diverging lens of focal length 50 cm

Solution

$$\begin{array}{l}\text { Ths is a combination of convex, concave and convex } \\\text { lens. Number them } 1,2,3 \\\qquad \begin{aligned}\text { given } f_{1} &=f_{3}=20 \mathrm{~cm} \\\mu_{1} &=\mu_{3}=1.5 \\\mu_{2} &=4/3\end{aligned}\end{array}$$
$$\begin{array}{l}\quad \mu_{2}=4 / 3 \\\text { Now densmaker formula } \frac{1}{f}=\left(\mu_{r}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\\text { for (1) } \quad \frac{1}{f_{1}}=(1.5-1)\left(\frac{1}{R_{1}}-\left(-\frac{1}{R}\right)\right)\end{array}$$$$\text { (equi convex lenS }\left.R_{1}=R_{2}\right)$$

$$\begin{array}{l}\frac{1}{20}=\frac{1}{2}\times \frac{2}{R} \\ R=20 \mathrm{~cm} \\\text { for (2) } \\\qquad \begin{array}{l}\frac{1}{f_{2}}=\left(\frac{4}{3}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \\\frac{1}{f_{2}}=\frac{1}{3}\left(\frac{-2}{R}\right) \\ f_{2}=-30\mathrm{~cm}\end{array}\end{array}$$
$$\begin{array}{l}\text { Now we know that } \\\qquad \begin{aligned}\text { ieffective } &=P_{1}+P_{2}+P_{3} \\\frac{1}{f_{e q}} &=\frac{1}{20}-\frac{1}{30}+\frac{1}{20} \\ f_{\text {eq }}=15\mathrm{~cm}\end{aligned} \\\text { So it is a converging lens of focal length } 15 \mathrm{~cm} .\end{array} $$
Question 4
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A ray incident at a point at an angle of incidence $$60$$$$^{\circ}$$ enters a glass sphere of $$\mu =\sqrt{3}$$ and is reflected and refracted at the further surface of the sphere. The angle between the reflected and refracted rays at this surface is :

A
$$50$$ $$^{\circ}$$

B
$$90$$$$^{\circ}$$

C
$$60$$$$^{\circ}$$

D
$$40$$$$^{\circ}$$

Solution

Using Snell's Law at point $$P$$ we get:-
$$\dfrac{sin\ 60^o}{sin\ r_1}=\sqrt3$$
$$sin \ r_1= \dfrac12\implies r_1= 30^o$$

As we know that $$r_1=r_2 \implies r_2=30^o$$

Using Snell's Law at point $$Q$$ we get:-

$$\dfrac{sin \ 30^o}{sin\ i_2}=\dfrac1{\sqrt3}\implies i_2= 60^o$$

Since reflection at point $$Q$$ occurs
$$\implies r_2'=r_2= 60^o$$

Let angle between the refracted ray and reflected ray be $$\alpha$$
$$\alpha= 180^o- (r_2'+r_2)= 90^o$$

Hence option $$(B)$$ is correct.
Question 5
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The difference in the number of wavelengths, when yellow light propagates through air and vacuum columns of the same thickness is one. The thickness of the air column is :
(Refractive index of air $$\mu _{a}=$$1.0003 ; Wavelength of yellow light in vacuum $$= 6000A^{\circ}$$)

A
$$1.8 mm$$

B
$$2 mm$$

C
$$2 cm$$

D
$$2.2 cm$$

Solution

No of wavelengths $$=\dfrac{thickness}{wavelength}$$
Let $$x$$ be the thickness of the air column

so $$\dfrac{x}{\lambda_{air}}-\dfrac{x}{\lambda _{vacuum}}=1$$

$$\dfrac{x}{\lambda }\left ( 1.0003-1 \right )=1$$

$$x=\dfrac{6000\times 10^{-10}}{0.0003}m$$

$$=2 mm$$
Question 6
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In the diagram shown

A
$$\mu ^{1}=\mu $$

B
$$\mu ^{1}<\mu $$

C
$$\mu ^{1}>\mu $$

D
$$\mu ^{1}\geq \mu $$

Solution

Since, the given lens is a convex lens hence it should converge a parallel beam of light falling on it to a single focal point. However, since it is showing the exactly opposite behaviour, thus it is definitely placed in a medium of higher refractive index.
Thus, we conclude that $$ {\mu}_{1} > {\mu} $$
Question 7
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In the case of refraction of light :
a) Frequency changes
b) Speed changes
c) Wavelength changes

A
a is correct

B
b and c are correct

C
a, b, c are correct

D
a and b are correct

Solution

In the process of refraction, speed of light and wavelength of light changes.

In a medium, speed of light is given by $$v=\dfrac{c}{\mu }$$ where $$\mu $$ is refractive index of the medium. Hence, for medium with different refractive index , speed of light is different.

Frequency will not change because at the boundary/interface of the medium, the number of waves you send is the number of waves you receive at the other side, almost instantly.

Since, for a travelling wave we have a relation $$ \nu = \dfrac{v}{\lambda} $$.
Hence wavelength also changes in the process of refraction.

So, option B is correct.
Question 8
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ACB is a right-angled glass prism with refractive index 1.5. Angle A, B, C are 60$$^{\circ}$$,30 $$^{\circ}$$and 90$$^{\circ}$$ respectively. A thin layer of liquid is on the AB. For a ray of light which is incident normally on AC to ;be totally reflected at AB, the refractive index of the liquid on AB should be

A
1.5

B
1.4

C
1.3

D
1.2

Solution

Given, $$\mu$$=refractive index of liquid.

From figure it is itself clear that , angle of incidence, $$i=60^{o}$$

Now, for internal reflection , angle of refraction $$r=90^{o}$$

Now, $$\dfrac{\sin i}{\sin r}=\dfrac{\mu}{\mu_1}$$

$$\implies \sin 60^{o}=\dfrac{\mu}{1.5}$$

$$\implies \mu=\dfrac{\sqrt{3}}{2}\times 1.5$$

$$\implies \mu=1.3$$

Hence, answer is option-(C)
Question 9
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A light ray is incident perpendicularly to one face of a 90$$^{\circ}$$ prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45$$^{\circ}$$, we conclude that the refractive index $$n$$ :

A
$$n>\dfrac{1}{\sqrt{2}}$$

B
$$n>\sqrt{2}$$

C
$$n<\dfrac{1}{\sqrt{2}}$$

D
$$n<\sqrt{2}$$

Solution

We know
$$\theta _{cric}=sin^{-1}(\dfrac{1}{n })$$

$$45^{o}> \theta $$

$$sin(45^o)> \dfrac{1}{n}$$

$$\Rightarrow n > \sqrt{2} $$
Question 10
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In the figure, light is incident on the thin lens as shown. The radius of curvature for both the surface is R. The focal length of this system is

A
$$f=\dfrac{\mu_3R}{\mu_3-\mu_1}$$

B
$$f=\dfrac{\mu_2R}{\mu_1-\mu_2}$$

C
$$f=\dfrac{\mu_3}{\mu_2-\mu_1}$$

D
$$f=\dfrac{\mu_1R}{\mu_1-\mu_2}$$

Solution

Consider refraction from the two spherical surfaces.
At surface 1,
$$\dfrac{\mu_2}{v_1}-\dfrac{\mu_1}{\infty}=\dfrac{\mu_2-\mu_1}{R}$$

At surface 2,
$$\dfrac{\mu_3}{v_2}-\dfrac{\mu_2}{v_1}=\dfrac{\mu_3-\mu_2}{R}$$

$$\implies v_2=\dfrac{\mu_3R}{\mu_3-\mu_1}=f$$