Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 38

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Question 1
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A ray of light from air suffers partial reflection and refraction at the boundary of water as shown in the figure. In Fig, which of the ray is the correct refracted ray?

A
$$A$$

B
$$B$$

C
$$C$$

D
$$D$$

Solution

When light travels from a rarer to denser medium, the refracted ray bends towards the normal.
According to Snell's Law: $$\dfrac{n_2}{n_1}=\dfrac{sin\theta_1}{sin\theta_2}$$
Refractive index is higher in a denser medium hence angle of refraction in denser medium should be lesser than angle of incidence. Thus, option B.
Question 2
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An object kept on the principal axis and infront of a spherical mirror, is moved along the axis itself. Its lateral magnification m is measured, and plotted versus object distance |u| for a range of u, as shown in fig. The magnification of the object when it is placed at a distance 20 cm in front of the mirror is:

A
$$-1$$

B
$$1$$

C
$$8$$

D
$$20$$

Solution

We need to find the focal length $$f$$ of the mirror. From graph we can notice that when object distance u is -5 cm then the lateral magnification is 2.
As magnification = $$-\dfrac{\text{Image distance}}{\text{Object distance}}$$.
So the image distance: $$v =-2\times 5=-10\ cm$$.
From the mirror equation, $$f=-10\ cm$$.
Now when $$u$$ is $$-20\ cm$$, using the mirror equation,
$$v = -20\ cm$$, or the image is at the center of curvature of the spherical mirror.
So the magnification here is $$=-\frac{(-20)}{(-20)}=-1$$
Question 3
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A ray of light is incident at the glass-water interface at an angle i. It merges finally parallel to the surface of water. Then, the value of $$\mu_g$$ would be :

A
$$(4/3) sin i$$

B
$$1/ sin i$$

C
$$4/3$$

D
$$1$$

Solution

Applying Snell's law for glass and water,
$$\mu_{g}sin\ i=\mu_{w}sin\ r . . . . . . . . . (i)$$

Applying Snell's law for water and air,
$$\mu_{w}sin\ r=\mu_{a}sin\ 90^0 . . . . . . (ii)$$

Using (i) and (ii), we can apply Snell's Law for glass and air

$$\mu_{g}sin\ i=\mu_{a}sin\ 90^0 $$
$$ \implies \mu_{g}sin\ i= 1 \times 1 $$

$$\implies \mu_{g}=\dfrac{1}{sin \ i}$$

Option $$B$$ is correct.
Question 4
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An object is kept at $$15\ cm$$ from a convex mirror of focal length $$25\ cm$$. What is the magnification?

A
$$4/9$$

B
$$5/8$$

C
$$9/4$$

D
$$8/5$$

Solution

Given, (Convex mirror)
Object distance, $$u=-15\ cm$$
Focal length, $$f=25\ cm$$
Image distance, $$v$$

Using mirror formula,
$$\dfrac1f=\dfrac1v+\dfrac1u$$

$$\dfrac1v=\dfrac1f-\dfrac1u$$

$$\dfrac{1}{v}=\dfrac{1}{25}-\dfrac{1}{-15}$$

$$\dfrac{1}{v}=\dfrac{3+5}{75}$$

$$v=\dfrac{75}{8}\ cm$$

Magnification, $$m=\dfrac{-v}{u}$$

$$m=\dfrac{-75}{8\times (-15)}$$

$$m=\dfrac58$$
Question 5
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Fig shows an object and its image formed by a thin lens. Then the nature and focal length of lens is

A
$$f=4.8$$ cm converging lens

B
$$f=-4.8$$ cm diverging lens

C
$$f=2.18$$ cm converging lens

D
$$f=-2.18$$ cm diverging lens

Solution

From figure, $$u = -(3+5) = -8$$ cm, $$v = -3$$ cm

Using lens formula, $$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$$

$$\therefore$$ $$\dfrac{1}{-3} - \dfrac{1}{-8} = \dfrac{1}{f}$$

$$\implies f = -4.8$$ cm

As the focal length of the lens comes out to be negative, thus its a diverging lens (concave lens).
Question 6
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The diagram alongside shows the refraction of a ray of light from air to a liquid. The refractive index of liquid with respect to air is :

A
$$\sqrt{\dfrac{1}{_{\displaystyle 2}}}$$

B
$$\sqrt{\dfrac{2}{_{\displaystyle 3}}}$$

C
$$\sqrt{\dfrac{3}{_{\displaystyle 2}}}$$

D
$$\sqrt{3}$$

Solution

According to Snell's law when light goes from rarer to denser medium ,
$$n_2 \sin r = n_1 \sin i$$ ...............$$(1)$$
where, $$n_2$$ is refractive index of denser medium(liquid), $$n_1$$ is refractive index of rarer medium(air).
Angle of incidence with the normal(dashed line in the figure ) = $$i$$
Angle of refraction with the normal = $$r$$
According to the given figure, $$ i = 60^o$$ and $$ r = 45^o$$
Define $$n = \dfrac {n_2}{n_1}$$ , which is refractive index of liquid with respect to air.
Using equation $$(1)$$ we get,
$$n= \dfrac{\sin 60^o}{\sin 45^o} =\dfrac{\dfrac{\sqrt3}{2}}{\dfrac{1}{\sqrt{2}}}$$
$$n = \sqrt{\dfrac{3}{2}}$$

Option C is correct.
Question 7
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A fish is looking at a $$1.0m$$ high plant of the edge of the pond. Will the plant appear shorter or taller than its actual height, to the fish

A
taller

B
shorter

C
no change

D
can't say

Solution

When light from the top of the plant strikes air-water surface it refracts and bends towards the normal and strikes the fish. To the fish the refracted ray is the one appearing to travel from the top of the plant and not the incident ray. When the refracted ray is extended backwards along a straight line, such that it was the original incident ray that didn't undergo refraction, it intersects at a point higher than the actual top of the plant. Hence, the plant appears taller.
Question 8
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In the given figure, name the ray which represents the correct path of light while emerging out through a glass blocks :

A
$$A$$

B
$$B$$

C
$$C$$

D
$$D$$

Solution

When travelling through the glass slab, the emergent ray is parallel to incident ray but laterally shifted which is represented by B. It undergoes refraction at first surface where it bends towards the normal at air-glass interface and at the second surface it bends away from the normal. Since, the surfaces are parallel, the incident and emergent rays are parallel.
Question 9
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The incident ray, the _______ ray and the normal lie in the same plane during refraction.

A
reflected

B
refracted

C
diffused

D
blocked

Solution

According to the first law of refraction, the incident ray, the refracted ray and the normal lie in the same plane.
Question 10
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Directions For Questions

Fill in the blank spaces by choosing the correct words from the list given below:
List: principal focus, optical center, pole, principal axis.

...view full instructions

A ray of light passing through the .......... of the lens, passes on undeviated.

A
principal focus

B
pole

C
optical centre

D
principal axis

Solution

Optical center
Any ray passing through the optical center will pass undeviated. Optical center is the center of the lens.