Ray Optics and Optical Test

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Ray Optics and Optical Test - 65

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Question 1
1 / -0

Light travels in two media A and B with speeds $1.8 \times { 10 }^{ 8 } m { s }^{ -1 }$ and $2.4 \times { 10 }^{ 8 } m { s }^{ -1 }$ respectively. Then the critical angle between them is

A
${ sin }^{ -1 }\left( \dfrac { 2 }{ 3 } \right)$

B
${ tan }^{ -1 }\left( \dfrac { 3 }{ 4 } \right)$

C
${ tan }^{ -1 }\left( \dfrac { 2 }{ 3 } \right)$

D
${ sin }^{ -1 }\left( \dfrac { 3 }{ 4 } \right)$

Solution

The Critical angle for the medium is given by:
$\Rightarrow\sin^{-1}\cfrac{1}{\mu}$

The refractive index of the medium A with respect to medium B is given by:
$^A\mu_{_B}=\cfrac{v_{2}}{v_{1}}$

$\mu=\cfrac{24}{18}=\cfrac{4}{3}$

Now the critical angle is obtained as:
$=\sin^{-1}\cfrac{3}{4}$
Question 2
1 / -0

Choose the correct answer from alternatives given.
The mixture of a pure liquid and a solution in a long vertical column (i.e, horizontal dimensions <<vertical dimensions) produces diffusion of solute particles and hence a refractive index gradient along the vertical dimension. A ray of light entering the column at right angles to the vertical is deviated from its original path. Find the deviation in travelling a horizontal distance d << h, the height of the column.

A
$- \, \dfrac{1}{d} \, \left(\frac{du}{dy} \right)\mu$

B
$\dfrac{d}{h} \, \left(\frac{du}{dy} \right)$

C
$\left(\frac{du}{dy} \right)h$

D
$- \, \dfrac{1}{\mu} \, \left(\frac{du}{dy} \right)d$

Solution

Let us consider a light ray entering the column at (x, y) at $90^o$ to the vertical. $\theta$ be the angle of incidence at x and it enter the thin column at height y. $d\theta$ is the deviation of ray between x and (x + dx) emerging at (x +dx, y + dy ) at angle $(\theta + d \theta)$ .
By snell's law,
$\mu (y) \, sin \, \theta \, =\, \mu (y +dy) \, sin \, ( \theta \, + \, d \theta )$
$= \, \left[\mu (y) + \dfrac{d\mu}{dy} dy \right] \, (sin \, \theta \, cos \, d \theta \, + \, cos \, \theta \, sin \, d \theta)$
As $d \theta \, is \, small, \, cos \, d \theta \, =\, 1 \, and \, sin \, d \theta \, \approx \, d\theta$
$\therefore \mu(y) \, cos \, \theta \, d \theta \, =\, \dfrac{d \mu}{dy} dy \, sin \, \theta ; \, d \theta = \dfrac{-1 }{\mu} \dfrac{d \mu dy \, tan \, \theta}{dy}$
$\left[ tan \, \theta \, = \, \dfrac{dx}{dy} and \, raplacing \, \mu (y) \, by \, \mu \right]$
$\theta = -\dfrac{1}{\mu} \dfrac{d \mu}{dy} \int_{o}^{d} dx = -\dfrac{1}{\mu} \left(\dfrac{d \mu}{dy}\right) d$
Question 3
1 / -0

A glass prism has a right-triangular cross-section ABC, with $\angle$ A $=90^o$ . A ray of light parallel to the hypotenuse BC and incident on the side AB emerges grazing the side AC. Another ray, again parallel to the hypotenuse BC, incident on the side AC suffers total internal reflection at the side AB. Find the range of the refractive index $\mu$ of the material of the prism?

A
$\sqrt{\dfrac{3}{2}} < \mu < \sqrt 2$

B
$\mu > \sqrt 3$

C
$\mu < \sqrt{\dfrac{3}{2}}$

D
$\sqrt 2 < \mu < \sqrt 3$

Solution

Since of is parallel to $B C$ .
at $A B$
(1) $\sin (90-\theta)=\mu \sin r$
$\frac{\cos \theta}{\mu}=\sin r$
at $A C$
$\begin{array}{l} \mu \sin (90-r)=(1) \sin 90 \\ \qquad \mu \cos r=1 \\ \sin ^{2} r+\cos ^{2}r=1 \\ \qquad \begin{array}{l} \frac{\cos ^{2} \theta}{\mu^{2}}+\frac{1}{\mu^{2}}=1 \\ \cos ^{2} \theta+1=\mu^{2}-(A) \end{array} \\ \qquad 2-\sin ^{2} \theta=\mu^{2} \Rightarrow \sin ^{2} \theta=2-\mu^{2}-(A) \end{array}$
at $\mathrm{AC}$
$(1) \sin \theta=\mu \sin r-0$
$\mu \sin (90-\gamma)>1$ ( at $A B$ condition for $T I R)$
$\cos x>\frac{1}{\mu} \longrightarrow 3$
By (1)
$\sin r=\frac{\sin \theta}{\mu} \Rightarrow \cos r=\frac{\sqrt{M^{2}-\sin ^{2} \theta}}{\mu}$
$\frac{\sqrt{\mu^{2}-\sin ^{2} \theta}}{\mu}>\frac{1}{\mu}$
$\mu^{2}>\sin ^{2} \theta+1$
By equation $A$
$\begin{array}{l}\mu^{2}>2\mu^{2}+1\\2\mu^{2}>3\\\mu>\sqrt{\frac{3}{2}}\end{array}$
for refraction to be happened At AC
$\sin \theta=\mu \sin r$
$\mu \sin r<1$
$\sin ^{2} x<\frac{1}{\mu^{2}}$
By equation (3) $\cos \gamma>\frac{1}{\mu}\Rightarrow 1-\sin ^{2} \gamma>\frac{1}{\mu^{2}}$
SO
$1-\frac{1}{\mu^{2}}<\frac{1}{\mu^{2}}$
$1-\frac{2}{\mu^{2}} < O\Rightarrow\mu<\sqrt{2}$
$\text { so }{\sqrt{\frac{3}{2}}}<\mu<\sqrt{2}$
So option (A) is correct
Question 4
1 / -0

A monochromatic light passes through a glass slab $\left (\mu = \dfrac {3}{2}\right )$ a thickness $90\ cm$ in time $t_{1}$ . If it takes $t_{2}$ to travel the same distance through water $\left (\mu = \dfrac {4}{3}\right )$ . The value of $(t_{1} - t_{2})$ is :

A
$5\times 10^{-8}s$

B
$5\times 10^{-10}s$

C
$2.5\times 10^{-10}s$

D
$2.5\times 10^{-8}s$

Solution

$C\to$ Speed of light in vacuum

$V \to$ Speed of light in medium

Refractive Index,

$\mu = C/V$

$time = distance/speed$

$\mu_1 = 3/2 (Glass)$

$V_1 = \dfrac{C_1}{\mu_1} = \dfrac{3\times 10^8}{3/2}= 2\times 10^8 m/s$

$d = 90cm = 0.9m$

$t_1 = d/V_1 = 0.9m / 2\times 10^8 m/s$

$t_1 = 0.45 \times 10^{-8}s$

$\mu_2 = 4/3 (water)$

$V_2 = \dfrac{C_1}{\mu_2} = \dfrac{3 \times 10^8 m/s}{4/3} = 2.25\times 10^8 m/s$

$d = 90cm = 0.9m$

$t_2 = \dfrac{d}{v_2} = \dfrac{0.9}{0.25\times 10^8m/s}$

$t_2 = 0.4\times 10^{-8}s$

$t_1-t_2 = 0.45\times 10^{-8}s - 0.40 \times 10^{-8}= 0.05\times 10^{-8} s$

$t_1-t_2 = 5\times10^{-10} \, s$
Question 5
1 / -0

A ray of light is incident on a thick slab of glass of thickness t as shown in the figure. The emergent ray is parallel to the incident ray but displaced sideways by a distance d. if the angles are small then d is

A
$t( 1-\dfrac { i }{ r })$

B
$rt( 1-\dfrac { i }{ r })$

C
$it( 1-\dfrac { r }{ i })$

D
$t( 1-\dfrac { r }{ i })$

Solution

The lateral shift occured in the ray is given by:
$d=\cfrac{t\sin(i-r)}{\cos r}$

For small angle, $\sin(i-r)\approx i-r$
$\cos r\approx 1$

$d=\cfrac{t(i-r)}{1}=it(1-\cfrac{r}{i})$
Question 6
1 / -0

Ocean appears blue due to _____________________.

A
Greater depth

B
Scattering of light by water particles

C
Blue colour of water

D
None of these

Solution

The ocean appears blue due to the scattering of light by water particles. When sunlight passes through the atmosphere, the fine particles in the air scatter the blue color (shorter wavelengths) more strongly than red.
Question 7
1 / -0

In the given figure, the radius of curvature of a curved surface for both the piano-convex and plano-concave lens is 10 cm and refractive index for both is 1.5. The location of the final image after all the refractions through lenses is:

A
$15 cm$

B
$20 cm$

C
$25 cm$

D
$40 cm$

Solution

The focal length of the plano - convex lens is
$\dfrac{1}{f} \, = \, (1.5 \, - \,1) \left ( \dfrac{1}{+10} \, - \, \dfrac{1}{\infty }\right ) \, = \, \dfrac{1}{20}$
Focal length of plano - concave lens is
$\dfrac{1}{f} \, = \, (1.5 \, - \,1) \left ( \dfrac{1}{\infty} \, - \, \dfrac{1}{10}\right ) \, = \, \dfrac{-1}{20}$
since parallel beams are incident on the lens, its image from plano - concave lens will be formed at $+20$ cm from it (at the focus) and will act as an object for the plano - concave lens. since the two lens are at a distance of $10$ cm from each other, therefore, for the next lens
$u = +10 cm.$
$\therefore v = \dfrac{uf}{u + f} = \dfrac{10 \, \times \, 20}{10 - 20}$
$= 20 cm$
Question 8
1 / -0

Choose the correct answer from alternatives given.
A ray of light travelling in a medium of refractive index $\mu$ is incident at an angle $\theta$ on a composite transparent plate consisting of 50 plates of RI. $1.0l\mu, \, 1.02\mu, \, 1.03\mu,\, ...., \, 1.50\mu.$ The ray emerges from the composite plate into a medium of refractive index $1 .6\mu$ at angle x. Then:

A
$sin \, x \, = \, \left(\dfrac{1.01}{1.5}\right)^{50} \, sin \, \theta$

B
$sin \, x \, = \, \dfrac{5}{8} \, sin \, \theta$

C
$sin \, x \, = \, \dfrac{8}{5} \, sin \, \theta$

D
$sin \, x \, = \, \left(\frac{1.5}{1.01}\right)^{50} \, sin \, \theta$

Solution

According to Snell's law, $\mu sin \theta \, = \, 1.6 \mu sin x$
$\therefore sin \, x = \, \dfrac{5}{8} \, sin \theta$
Question 9
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A point object O is placed at a distance of 20 cm in front of a equiconvex lens $(^a\mu_g \, = \, 1.5)$ of focal length 10 cm. The lens is placed on a liquid of refractive index 2 as shown in the figure. An image will be formed at a distance h from a lens. The value of h is

A
5 cm

B
10 cm

C
20 cm

D
40 cm

Solution

As $\dfrac{1}{f} = (\mu - 1) \left[\dfrac{1}{R_1} - \dfrac{1}{R_2}\right]$
$\dfrac{1}{10} = (1.5 - 1) \left(\dfrac{1}{R} + \dfrac{1}{R} \right) = 0.5 \times \dfrac{2}{R} \Rightarrow R = 10 \, cm$
$\therefore \dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}$
Refraction from first surface,
$\dfrac{1.5}{v_1} - \dfrac{1}{-20} = \dfrac{1.5 - 1}{+10} \Rightarrow v_1 = \infty$
For the second surface,
$\dfrac{2}{v} - \dfrac{1.5}{\infty} = \dfrac{2 - 1.5}{-10} \Rightarrow v = -40 \, cm \,\, \therefore h = 40 \, cm$
Question 10
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A ray incident at a point at an angle of incidence of ${60}^{o}$ enters a glass sphere of $R.l.n=\sqrt 3$ and is reflected and refracted at the further surface of the sphere. The angel between the reflected and refracted rays at this surface is

A
${50}^{o}$

B
${60}^{o}$

C
${90}^{o}$

D
${40}^{o}$

Solution

Refraction at $P$ , $\dfrac { \sin { 60 }^{o} }{ \sin { { r }_{ 1 } } } =\sqrt { 3 }$
$\Rightarrow \sin { { r }_{ 1 } } =1/2 \quad \Rightarrow { r }_{ 1 }=30$
Since ${ { r }_{ 2 } } ={ r }_{ 1 } \quad \therefore \quad{ r }_{ 2 }=30$
Reflection at $Q$ , $\dfrac { \sin { { r }_{ 2 } } }{ \sin { { i }_{ 2 } } } =\dfrac { 1 }{ \sqrt { 3 } }$
Putting ${R}_{2}={30}^{o}$ we obtain ${i}_{2}={60}^{o}$
Reflection at $Q, {r}^{\prime }_{2}={r}_{2}$
$\therefore \quad \alpha =180^{o}-({ r }^{ \prime }_{ 2 }+{ r }_{ 2 })=180^{o}-(30^{o}+60^{o})=90^{o}$
Hence (C) is correct.

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Re-Attempt Weekly Quiz Competition

Ray Optics and Optical Test - 65

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Question 1

sin−1(23){ sin }^{ -1 }\left( \dfrac { 2 }{ 3 } \right) sin−1(32​)

tan−1(34){ tan }^{ -1 }\left( \dfrac { 3 }{ 4 } \right)tan−1(43​)

tan−1(23){ tan }^{ -1 }\left( \dfrac { 2 }{ 3 } \right)tan−1(32​)

sin−1(34){ sin }^{ -1 }\left( \dfrac { 3 }{ 4 } \right)sin−1(43​)

Solution

Question 2

− 1d (dudy)μ- \, \dfrac{1}{d} \, \left(\frac{du}{dy} \right)\mu−d1​(dydu​)μ

dh (dudy) \dfrac{d}{h} \, \left(\frac{du}{dy} \right)hd​(dydu​)

(dudy)h\left(\frac{du}{dy} \right)h(dydu​)h

− 1μ (dudy)d- \, \dfrac{1}{\mu} \, \left(\frac{du}{dy} \right)d−μ1​(dydu​)d

Solution

Question 3

32<μ<2\sqrt{\dfrac{3}{2}} < \mu < \sqrt 223​​<μ<2​

μ>3\mu > \sqrt 3μ>3​

μ<32\mu < \sqrt{\dfrac{3}{2}}μ<23​​

2<μ<3\sqrt 2 < \mu < \sqrt 32​<μ<3​

Solution

Question 4

5×10−8s5\times 10^{-8}s5×10−8s

5×10−10s5\times 10^{-10}s5×10−10s

2.5×10−10s2.5\times 10^{-10}s2.5×10−10s

2.5×10−8s2.5\times 10^{-8}s2.5×10−8s

Solution

Question 5

t(1−ir)t( 1-\dfrac { i }{ r })t(1−ri​)

rt(1−ir)rt( 1-\dfrac { i }{ r })rt(1−ri​)

it(1−ri)it( 1-\dfrac { r }{ i })it(1−ir​)

t(1−ri)t( 1-\dfrac { r }{ i })t(1−ir​)

Solution

Question 6

Greater depth

Scattering of light by water particles

Blue colour of water

None of these

Solution

Question 7

15cm15 cm15cm

20cm20 cm20cm

25cm25 cm25cm

40cm40 cm40cm

Solution

Question 8

sin x = (1.011.5)50 sin θsin \, x \, = \, \left(\dfrac{1.01}{1.5}\right)^{50} \, sin \, \thetasinx=(1.51.01​)50sinθ

sin x = 58 sin θsin \, x \, = \, \dfrac{5}{8} \, sin \, \thetasinx=85​sinθ

sin x = 85 sin θsin \, x \, = \, \dfrac{8}{5} \, sin \, \thetasinx=58​sinθ

sin x = (1.51.01)50 sin θsin \, x \, = \, \left(\frac{1.5}{1.01}\right)^{50} \, sin \, \thetasinx=(1.011.5​)50sinθ

Solution

Question 9

5 cm

10 cm

20 cm

40 cm

Solution

Question 10

50o{50}^{o}50o

60o{60}^{o}60o

90o{90}^{o}90o

40o{40}^{o}40o

Solution

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${ sin }^{ -1 }\left( \dfrac { 2 }{ 3 } \right)$

${ tan }^{ -1 }\left( \dfrac { 3 }{ 4 } \right)$

${ tan }^{ -1 }\left( \dfrac { 2 }{ 3 } \right)$

${ sin }^{ -1 }\left( \dfrac { 3 }{ 4 } \right)$

$- \, \dfrac{1}{d} \, \left(\frac{du}{dy} \right)\mu$

$\dfrac{d}{h} \, \left(\frac{du}{dy} \right)$

$\left(\frac{du}{dy} \right)h$

$- \, \dfrac{1}{\mu} \, \left(\frac{du}{dy} \right)d$

$\sqrt{\dfrac{3}{2}} < \mu < \sqrt 2$

$\mu > \sqrt 3$

$\mu < \sqrt{\dfrac{3}{2}}$

$\sqrt 2 < \mu < \sqrt 3$

$5\times 10^{-8}s$

$5\times 10^{-10}s$

$2.5\times 10^{-10}s$

$2.5\times 10^{-8}s$

$t( 1-\dfrac { i }{ r })$

$rt( 1-\dfrac { i }{ r })$

$it( 1-\dfrac { r }{ i })$

$t( 1-\dfrac { r }{ i })$

$15 cm$

$20 cm$

$25 cm$

$40 cm$

$sin \, x \, = \, \left(\dfrac{1.01}{1.5}\right)^{50} \, sin \, \theta$

$sin \, x \, = \, \dfrac{5}{8} \, sin \, \theta$

$sin \, x \, = \, \dfrac{8}{5} \, sin \, \theta$

$sin \, x \, = \, \left(\frac{1.5}{1.01}\right)^{50} \, sin \, \theta$

${50}^{o}$

${60}^{o}$

${90}^{o}$

${40}^{o}$