Light Reflection and Refraction Test

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Light Reflection and Refraction Test - 70

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Weekly Quiz Competition

Question 1
1 / -0

Curved mirror whose reflecting surface faces towards the centre of the sphere, from which is considered to be a part of, is known as

A
Convex mirror

B
Concave mirror

C
Plane mirror

D
None

Solution

The given picture shows the mirror of aforesaid description and this is a concave mirror.
Question 2
1 / -0

Which of the following is a spherical lens?

A
Concave lens

B
Convex lens

C
Diverging lens

D
All

Solution

All of the above lenses have one or both surface spherical and hence are spherical lenses.

Answer-(D).
Question 3
1 / -0

A beam of light striking the surface of a glass plate from the air as shown in above figure.
If the angle between reflected beam and refracted beam is 90 degree, Find out the refractive index of the glass?

A
$$\sin{{55}^{o}}$$

B
$$\dfrac{1}{\sin{{55}^{o}}}$$

C
$$\dfrac{1}{\sin{{35}^{o}}}$$

D
$$\dfrac{\sin{{55}^{o}}}{\sin{{35}^{o}}}$$

E
$$\dfrac{\sin{{35}^{o}}}{\sin{{55}^{o}}}$$

Solution

According to figure, $$\theta_1 + 35^o =90^o$$ $$\implies \theta_1 = 55^o$$
As the angle of incidence is equal to angle of reflection, thus $$\theta_2 =\theta_1 = 55^o$$
Also $$\theta_3 = 90^o - \theta_2 = 90^o - 55^o =35^o$$
Given : $$\theta_3 + \theta_4 = 90^o$$
$$\therefore$$ $$\theta_4 =90^o - 35^o = 55^o$$
$$\implies$$ $$\theta_5 =90^o - \theta_4 = 90^o - 55^o = 35^o$$
Using Snell's law of refraction : $$n_a \times sin \theta_1 = n_g \times sin\theta_5$$ where $$n_a = 1$$
$$\therefore$$ $$1 \times sin55^o =n_g \times sin35^o$$ $$\implies n_g = \dfrac{sin55^o}{sin 35^o}$$
Question 4
1 / -0

A glass slab has two long parallel faces. Light ray is incident on the glass slab and trace of light ray is drawn. Now glass slab is shifted parallel to the parallel faces. What will be the effect on the trace of light?

A
It will shift laterally

B
Angle of deviation will change

C
No effect

D
Depends on amount of shift

Solution

Since thickness, angle of incidence and refractive index is same, there is no change in path of light ray.
Question 5
1 / -0

A ray of light is incident on the interface between water and glass at an angle i and refracted parallel to the water surface, then value of $$\mu_g$$, refractive index of glass will be:

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

B
$$\displaystyle\frac{1}{\sin i}$$

C
$$\displaystyle\frac{4}{3}$$

D
$$1$$

Solution

Applying Snell's Law at glass-water interface,
$$\mu_g \sin i=\dfrac{4}{3}\sin r$$
Applying Snell's Law at water-air surface,
The angle of refraction at glass-water interface will be same as angle of incidence at air-water interface.
$$\dfrac{4}{3}\sin r=1\sin 90^{\circ}=1$$
$$\implies \mu_g \sin i=1$$
$$\implies \mu_g=\dfrac{1}{\sin i}$$
Question 6
1 / -0

A spherical lens has a focal length $$2\ cm$$. The lens will be

A
Concave lens

B
Convex lens

C
Concavo convex lens

D
None

Solution

Concave lens has negative and concavo-convex lens can have negative focal length.
But, convex lens always has positive focal length.

Answer-(B).
Question 7
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The diameter of the reflecting surface of spherical mirror is called ..............

A
Focus

B
Aperture

C
Centre of curvature

D
None of these

Solution

The diameter of the reflecting surface of the spherical mirror is called it's $$aperture$$.
Question 8
1 / -0

Magnification of lens is:

A
$$\dfrac {\text {Size of the image}}{\text {Size of the object}}$$

B
$$\dfrac {\text {Image distance}}{\text {Object distance}}$$

C
Both A and B

D
None

Solution

For lens, magnification is given by
$$m=\dfrac{h_i}{h_o}=\dfrac{v}{u}$$
where $$h_i$$ = image height $$h_0$$ = object height $$v$$=image distance $$u$$ = object distance

Answer-(c)
Question 9
1 / -0

The radius of the sphere of which the reflecting surface of a spherical mirror forms a part, is called ...............

A
Centre of curvature

B
Radius of curvature

C
Focus

D
Pole

Solution

The radius of the sphere of which the reflecting surface of a spherical mirror forms a part, is called the radius of curvature of the mirror. Here, the distance $$R$$ is the radius of curvature in the figure.

Answer-(B).
Question 10
1 / -0

A water film is formed on a glass-block. A light ray is incident on water film from air at an angle of $$\displaystyle { 60 }^{ \circ }$$ with the normal. The angle of incidence on glass slab is
($$\displaystyle { \mu }_{ g } = 1.5,{ \mu }_{ w } = \frac { 4 }{ 3 } $$)

A
$$\displaystyle { sin }^{ -1 }\left( \frac { 3\sqrt { 3 } }{ 8 } \right) $$

B
$$\displaystyle { sin }^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right) $$

C
$$\displaystyle { sin }^{ -1 }\left( \frac { 4\sqrt { 3 } }{ 9 } \right) $$

D
$$\displaystyle { sin }^{ -1 }\left( \frac { 9\sqrt { 3 } }{ 16 } \right) $$

Solution

The angle on incidence on glass slab will be equal to the angle of refraction on the air water interface.

Applying Snell's Law on air water interface, we get
$$\mu_asin \ i = \mu_wsin \ r$$

$$1 \times sin \ { 60 }^{ 0 }=\dfrac { 4 }{ 3 } sin \ r$$

$$sin \ r =\dfrac { 3\sqrt { 3 } }{ 8 } $$

$$r ={ sin }^{ -1 } \dfrac { 3\sqrt { 3 } }{ 8 } $$