Light Reflection and Refraction Test

Result

Light Reflection and Refraction Test - 32

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Question 1
1 / -0

Statement- I : The formula connecting $$u$$ , $$v$$ and $$f$$ for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature.
Statement- II : Laws of reflection are strictly valid for plane surfaces, but not for large spherical surfaces

A
Statement -I is true, Statement -II is true; Statement -II is a correct explanation for statement -I

B
Statement-I is true, Statement -II is true; Statement-II is not a correct explanation for statement- I

C
Statement - I is true, Statement -II is false

D
Statement - I is false, Statement -II is true

Solution

During the derivation of the mirror formula, we have to take few approximations for getting the mirror equation and those approximations include mirror must have very small size in comparison to the radius of curvature.
However, the laws of reflection are valid for any kind of reflecting surface.

Hence, statement - I is True, but statement - II is False.
Question 2
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The line NO that is perpendicular to the plane mirror surface and drawn at the point of incidence is called the :

A
reflected ray

B
incident ray

C
normal

D
refracted ray

Solution

The line NO that is perpendicular to the plane mirror surface and drawn at the point of incidence is called the normal.
Question 3
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A light ray is traveling from air to glass. The angle of incidence on the boundary is 30$$^{\circ}$$. Find the sine of the angle of refraction. (Take: $$sin\ 30^o = \dfrac{1}{2}$$)

A
1/3

B
2

C
3

D
1

Solution

Given:
The angle of incidence is $$i=30^o$$
Refractive index of glass is $$n_g=1.5$$
Refractive index of air is $$n_a=1$$
The sine angle of refraction $$sin\ r=?$$

using Snell's law we can write, $$n_a\times sin\ i = n_g \times sin\ r$$
$$\therefore sin\ r = \dfrac{n_a\times sin\ i}{n_g}=\dfrac{1\times sin\ 3o^o}{1.5}= \dfrac{1}{3}$$
Question 4
1 / -0

The angle of reflection is given by the angle :

A
NOQ

B
QOB

C
PON

D
AOP

Solution

The angle between the reflected ray and the normal drawn at the point of incidence to a reflecting surface is known as the angle of reflection.

In the given figure, angle NOQ is the angle of reflection.
Question 5
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Beams of light are incident through the holes A and B and emerge out of box through the holes C and D respectively as shown in the figure. Which of the following could be inside the box?

A
A rectangular glass slab

B
A convex lens

C
A concave lens

D
A prism

Solution

Lateral displacement is the distance by which the incident light has been displaced after bending through the device. Among the options, only glass slab has the property of laterally shifting an incident ray of light.
Question 6
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The point C is the :

A
Center of curvature

B
Principal focus

C
Pole

D
None

Solution

Point C is called the centre of curvature.
Question 7
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Magnification produced by a rear view mirror fitted in vehicles:

A
is less than one

B
is more than one

C
is equal to one

D
can be more than or less than one depending upon the position of the object in front of it

Solution
Question 8
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Which of the following defines the center of curvature of a curved mirror?

A
The center of the hollow glass sphere of which the curved mirror was (previously) a part

B
The geometric center of the curved mirror

C
The radius of the hollow glass sphere of which the curved mirror was (previously) a part

D
None of these

Solution

The center of curvature is the center of the hollow glass sphere of which the curved mirror was (previously) a part.
Question 9
1 / -0

The mirror shown is _______ in shape.

A
Convex

B
Concave

C
Concavo-convex

D
Convexo-concave

Solution

Given mirror is know as concave mirror because its reflecting surface is concave.
Question 10
1 / -0

The given figure shows a ray of light as it travels from medium A to medium B. Refractive index of the medium B relative to medium A is:

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

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

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

D
$$\sqrt{2}$$

Solution

Answer is A.

The Snell's law simply relates angles $$i$$ and $$r$$ to the refraction indices of the two media A and B. Therefore, refractive index of medium B relative to medium A is:
$$n=\dfrac { sin\quad i }{ sin\quad r }=\dfrac{sin\ 60^0}{sin\ 45^0} =\dfrac { \sqrt { 3 } }{ \sqrt { 2 } } $$.