Ray Optics and Optical Test

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Ray Optics and Optical Test - 59

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Weekly Quiz Competition

Question 1
1 / -0

The focal length of a concave mirror is $$50\ cm.$$ Where an object be placed so that its image is two times and inverted

A
$$75\ cm$$

B
$$72\ cm$$

C
$$63\ cm$$

D
$$50\ cm$$

Solution

for Real image $$m=-2,$$ so by using $$m=\dfrac{f}{f-u}$$
$$\Rightarrow -2=\dfrac{-50}{-50-u}\Rightarrow u=-75\ cm$$
Question 2
1 / -0

The critical angle between an equilateral prism and air is $$45^{o}$$. If the incident ray is perpendicular to the refracting surface, then

A
After deviation it will emerge from the second refracting surface

B
It is totally reflected on the second and third emerges out perpendicularly from third surface in air

C
It is totally reflected on the second and third refracting surface and finally emerges out from the first surface

D
It is a totally reflected from all the three sides of prism and never emerges out

Solution

From the above figure,
It is totally reflected on the second and third emerges out perpendicularly from the third surface in air.
Question 3
1 / -0

Two parallel pillars are $$11\ km$$ away from an observer. The minimum distance between the pillars so that they can be seen separately will be

A
$$3.2\ m$$

B
$$20.8\ m$$

C
$$91.5\ m$$

D
$$183\ m$$

Solution

As limit of resolution of eye is $$\left( \dfrac{1}{60}\right)^o$$, the pillars will be seen
distinctly if $$\theta > \left( \dfrac{1}{60}\right)^o$$
i.e. $$\dfrac dx > \left( \dfrac{1}{60}\right)\times \dfrac{\pi}{180}$$
$$\Rightarrow d > \dfrac{\pi \times x}{60\times 180}$$
$$\Rightarrow d > \dfrac{3.14 \times 11 \times 10^3}{60\times 180}\Rightarrow d > 3.2\ m$$
Question 4
1 / -0

If the speed of light in vacuum is $$C\ \ m/sec$$, then the velocity of light in a medium of refractive index $$1.5$$ is

A
$$1.5\times C$$

B
$$C$$

C
$$\dfrac {C}{1.5}$$

D
Can have any velocity

Solution

Spped of light in vacuum $$=C$$ m/s
Refrective index of medium $$\mu=1.5$$
Velocity of light in any medium having refrective index $$\mu$$, $$v=\dfrac C\mu$$
On putting the value of $$\mu$$, $$v=\dfrac C{1.5}$$

Option D.
Question 5
1 / -0

On a glass plate a light wave is incident at an angle of $$60^o$$. If the reflected and the refracted waves are mutually perpendicular, the refractive index of material is

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

B
$$\sqrt 3$$

C
$$\dfrac {3}{2}$$

D
$$\dfrac {1}{\sqrt 3}$$

Solution

From figure
$$ < i=60^o, < r=30^o$$
so $$\mu =\dfrac {\sin 60}{\sin 30}=\sqrt 3$$
Question 6
1 / -0

A ray of light strikes a plane mirror $$M$$ at an angle of $$45^o$$ as shown in the figure. After reflection, the ray passes through a prism of refractive index $4\$1.5$$ whose apex angle is $$4^o$$. The total angle through which the ray is deviated is

A
$$90^o$$

B
$$91^o$$

C
$$92^o$$

D
$$93^o$$

Solution

As we know,
$$\delta_{net}=\delta_{mirror}+\delta_{prism}$$
$$=(180-2i)+(\mu-1)A$$
$$=(180-2\times 45)+(1.5-1)\times 4=92^{o}$$
Question 7
1 / -0

The image distance of an object placed $$10\ cm%$$ in front of a thin lens of focal length $$5cm$$ is

A
$$6.5\ cm$$

B
$$8.0\ cm$$

C
$$9.5\ cm$$

D
$$10.0\ cm$$

Solution

As we know,
$$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \Rightarrow \dfrac{1}{+5}=\dfrac{1}{v}-\dfrac{1}{(-10)} \Rightarrow v=10\ cm$$
Question 8
1 / -0

A prism $$(\mu=1.5)$$ has the refracting angle of $$30^{o}$$. The deviation of a monochromatic ray incident normally on its one surface will be $$(\sin 48^{o}36=0.75)$$

A
$$18^{o}36'$$

B
$$20^{o}30'$$

C
$$18^{o}$$

D
$$22^{o}1'$$

Solution

For surface $$AC\dfrac{1}{\mu}=\dfrac{\sin 30^{o}}{\sin e}\Rightarrow \sin e=\mu\sin 30^{o}$$
$$\Rightarrow \sin e=1.5\times \dfrac{1}{2}=0.75$$
$$\Rightarrow e=\sin^{-1}(0.75)=48^{o}36'$$
From figure $$\delta =e-30^{o}$$
$$=48^{o}36'-30^{o}=18^{o}36'$$
Question 9
1 / -0

Absolute refractive indices of glass and water are $$\dfrac {3}{2}$$ and $$\dfrac {4}{3}$$. The ratio of velocity of light in glass and water will be

A
$$4:3$$

B
$$8:7$$

C
$$8:9$$

D
$$3:4$$

Solution

We know that , $$\mu = \dfrac{c}{v} \Rightarrow v=\dfrac{c}{\mu}$$

Since, $$c$$ is constant, we can say that

$$v\propto \dfrac {1}{\mu}\Rightarrow \dfrac {v_1}{v_2}=\dfrac {\mu_2}{\mu_1}$$

Given that, $$\mu_g = \dfrac{3}{2}$$ and $$\mu_w = \dfrac{4}{3}$$

$$\Rightarrow \dfrac {v_g}{v_w}=\dfrac {\mu_w}{\mu_g}=\dfrac {4/3}{3/2}=\dfrac {8}{9}$$
Question 10
1 / -0

The graph between the lateral magnification $$(m)$$ produced by a lens and the distance of the image $$(v)$$ is given by

A

B

C

D

Solution

For a lens $$m=\dfrac{f-v}{f}\Rightarrow m=\left(-\dfrac{1}{f}\right)v+1$$
Comparing this equation with $$y = mx + c$$ (equation of straight line)
Hence, graph will be a straight line with negative slope.