Light Reflection and Refraction Test

Result

Light Reflection and Refraction Test - 41

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Focus is real in the case of

A
Concave mirror

B
Convex mirror

C
Plane mirror

D
None of the above

Solution

Focus is real in the case of a concave mirror because its reflecting surface is curved inwards and the parallel rays converge at the point.
Question 2
1 / -0

In a spherical mirror, normal drawn on any point on a spherical mirror passes through :

A
Focus

B
Pole of the mirror

C
Centre of curvature

D
None of the above

Solution

In a spherical mirror, normal drawn on any point on a spherical mirror passes through the centre of curvature.

Question 3

1 / -0

The first law of refraction states that the incident and refracted rays are in the same ________ as the normal.

phase

plane

both

none

Solution

Correct answer is (B) plane

The first law of refraction states that the incident and refracted rays are in the same plane as the normal.

The normal line is to the surface of the mirror which makes a 90 degree angle to the mirror. The angle of incidence is the angle that the incident ray makes the normal and the angle of reflected angle that the reflected ray makes with the normal.

Question 4
1 / -0

When the distance between the object and the plane mirror increases:

A
the image distance remains same

B
the size of the image will become less than the size of the object

C
the distance between the image and the plane mirror increases

D
the distance between the image and the plane mirror decreases

Solution

For a plane mirror, object distance $$=$$ image distance.
So, if the distance between the object and the plane mirror increases, the distance between the image and the plane mirror also increases.
Question 5
1 / -0

Focus is virtual in the case of

A
Concave mirror

B
Convex mirror

C
Plane mirror

D
None of the above

Solution

Focus is virtual in the case of convex mirror because the reflecting surface is curved outwards and the parallel rays appear to come from or diverge from this point. It exists virtually.
Question 6
1 / -0

As shown in the figure, the point $$C$$ is called the

A
optical center of the lens.

B
principal axis of the lens.

C
focus of the lens.

D
power of the lens.

Solution

The point $$C$$ is called as the optical center of the lens. It is a point on the principal axis of a lens through which light passes without undergoing any deviation.
Question 7
1 / -0

The second law of refraction states that :

A
$$\dfrac{\sin^2 i}{\sin^2 r}=\mu$$

B
$$\dfrac{\sin i}{\sin r}=\mu$$

C
$$\sqrt{\dfrac{\sin i}{\sin r}}=\mu$$

D
none of these are true

Solution

According to second Law of refraction, When incident ray, coming from one medium to the boundary of another medium, gets refracted, the ratio of the sine of angle of incidence i to the sine of angle of refraction r is always constant for a given wavelength of light.

This constant is the refractive index of second medium relative to the first medium.

This can be written as:
$$\dfrac{sin \ i}{sin \ r}= \ constant (\mu)$$
This is also known as Snell's law.
Question 8
1 / -0

If the refractive index of medium $$b$$ with respect to medium $$a$$ and that of medium $$c$$ with respect to medium $$b$$ are given as $$n_{ba}$$ and $$n_{cb}$$ respectively, then the refractive index of a medium $$c$$ with respect to a medium $$a$$ can be found by using the relation
$$n_{ca}\, =\, n_{ba}\, \times\, ........$$

A
$$n_{cb}$$

B
$$n_{ac}$$

C
$$n_{ca}$$

D
$$n_{ba}$$

Solution

The refractive index of medium $$c$$ with respect to $$a$$ is denoted as $$n_{ca}$$.
As per the definition, $$n_{ca} = \dfrac{n_a}{n_c}$$, where $$n_c$$ is the absolute refractive index of medium $$c$$ and $$n_a$$ is the absolute refractive index of medium $$a$$.
Now, $$n_{ca} = \dfrac{n_a}{n_c} = \dfrac{n_a}{n_b}\times \dfrac{n_b}{n_c}$$
$$\dfrac { { n }_{ a } }{ { n }_{ b } } ={ n }_{ ba }$$ and $$ \dfrac { { n }_{ b } }{ { n }_{ c } } ={ n }_{ cb }$$

Thus we get,
$${ n }_{ ba }\times { n }_{ cb } ={ n }_{ ca }$$
So the answer is $${ n }_{ cb }$$.
Question 9
1 / -0

A coin kept in a container and not visible can be viewed by pouring water into the container. It happens because of the

A
Reflection of light.

B
Refraction of light.

C
Variable refractive index of water.

D
None of the above.

Solution

When water is poured then it is due to refraction of light the coin becomes visible. When the light rays travel from the water medium (denser) to the air medium (rarer) it bends away from the normal due to the refraction. Therefore, an image of the coin is formed at a smaller depth causing it to be visible (as shown in the figure).
Question 10
1 / -0

The angle between the normal and refracted ray is called :

A
angle of deviation

B
angle of incidence

C
angle of refraction

D
none of these

Solution

Correct answer is (c)

The angle formed between the normal and refracted ray at the point of refraction is called angle of refraction.