Light - Reflection Refraction(10th Grade > Physics ) Questions and answers for exam Preparation

10th Grade > Physics

LIGHT - REFLECTION REFRACTION MCQs

Total Questions : 39 | Page 3 of 4 pages

A plane mirror and an object approach each other with speeds of $5 m s^{- 1}$ and $10 m s^{- 1}$ respectively. What will be the speed of the image w.r.t. the stationary observer?

$5 m s^{- 1}$
$15 m s^{- 1}$
$20 m s^{- 1}$
$25 m s^{- 1}$

Discuss Question

Answer: Option D. ->

25 m s^{- 1}

:
D

Given:
Speed of object, $u_{o} = 5 m s^{- 1}$
Speed of mirror, $u_{m} = 10 m s^{- 1}$
Mirror and object are approaching each other.
To find:
Speed of image
Procedure:
Let initial position of object be at $x_{o i} = 0$
Let initial position of mirror be at $x_{m i} = u$
Object distance equals the image distance. Hence, initial position of image is $x_{i i} = 2 u$
Distance is the product of speed and time.
After time $t$ ,
Position of object will be:
$x_{o f} = x_{o i} + u_{o} t = 5 t$
Position of mirror will be:
$x_{m f} = x_{m i} - u_{f} t = u - 10 t$
Let the final position of image be $x_{i f}$
Using object distance equals image distance after time $t$ .
$x_{i f} - x_{m f} = x_{m f} - x_{o f}$
$x_{i f} = 2 u - 25 t$
Distance moved by image in time $t$ is:
$s_{i} = x_{i f} - x_{i i}$
$s_{i} = - 25 t$
Speed is the ratio of distance and time. Speed of the image is:
$v_{i} = \frac{s_{i}}{t} = - 25$
Negative sign appears because image is moving leftwards.
Hence, speed of image is $25 m s^{- 1}$

Question 22.

In the mirror shown, the reflected ray will ___.

pass through F
pass through C
appear to come from F
appear to come from C

Discuss Question

Answer: Option C. -> appear to come from F
:
C

Ray parallel to principal axis after hitting the convex mirror will appear to come from its focus.

Question 23.

Observe the optical systems above. Given that all surfaces are made of transparent glass and the dotted surfaces represent plane surface. Which among them can be considered as lenses?

I, II, IV, V
I, II, III, V
I, II, III, IV
I, IV only

Discuss Question

Answer: Option C. -> I, II, III, IV
:
C

Lens should have at least one curved surface, so that it can converge or diverge the incoming light rays. The first optical system is called convex lens and the second one is concave lens. The third lens is called a plano-concave lens and the fourth one is called a plano-convex lens. The fifth optical system has two plane surfaces, and hence it can't be considered as lens.

Question 24.

An object is placed in front of a convex mirror at a distance of $60 m$ from its pole. Object starts moving towards the mirror at $20 m s^{- 1}$ . Find the average speed of the image for a time interval of $2 s$ if the focal length of the mirror is $20 m$ .

$0.568 m s^{- 1}$
$1 m s^{- 1}$
$3 m s^{- 1}$
$2.5 m s^{- 1}$

Discuss Question

Answer: Option D. ->

2.5 m s^{- 1}

:
D

Given:
Initial object distance, $u = - 60 m$
Speed of object, $v_{o} = 20 m s^{- 1}$
Focal length, $f = 20 m$
Time interval, $t = 2 s$
Applying mirror formula to find the initial position of the image $v$ :
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} + \frac{1}{- 60} = \frac{1}{20}$
$v = 15 m$

Distance travelled by the object is the product of speed and time.
$s_{o} = v_{o} t$
$s_{o} = 20 \times 2 = 40 m$
Therefore, now the object distance:
$u_{f} = u + s_{o} = - 60 + 40 = - 20 m$
Applying mirror formula to find the final position of the image $v_{f}$ :
$\frac{1}{v_{f}} + \frac{1}{u_{f}} = \frac{1}{f}$
$\frac{1}{v_{f}} + \frac{1}{- 20} = \frac{1}{20}$
$v_{f} = 10 m$
Average speed is the ratio of total distance travelled to the total time taken.
Average speed of the image:
$v_{i} = \frac{v - v_{f}}{t}$
$v_{i} = \frac{15 - 10}{2} = 2.5 m s^{- 1}$

Question 25.

An object is placed at a distance of $30 c m$ from the lens. Its image is formed at a distance of $10 c m$ on the same side of the lens. Find the focal length of the lens used.

$f = 10 c m$
$f = - 10 c m$
$f = + 15 c m$
$f = - 15 c m$

Discuss Question

Answer: Option D. ->

f = - 15 c m

:
D
Given:
Image distance

v = - 10 c m

Object distance

u = - 30 c m

(Using the sign conventions)
Let focal length of the lens be

f

.
Using the lens formula,

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

\frac{1}{f} = (\frac{1}{- 10}) - (\frac{1}{- 30})

f = - 15 c m

Here, negative sign for focal length shows that the lens is concave.

Question 26.

An object is placed at $60 m$ in front of a concave mirror. If its radius of curvature is $40 m$ , then find the image distance.

$20 m$
$30 m$
$60 m$
$120 m$

Discuss Question

Answer: Option B. ->

30 m

:
B

Given:
Radius of curvature, $R = - 40 m$
Object distance, $u = - 60 m$
Let the sign convention be such that the object distance is negative i.e. positive direction is taken along the direction of the incident ray.
From mirror formula,
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
where:
$v :$ Position of image
$f :$ Focal length
But focal length is half of the radius.
$⟹ f = \frac{R}{2} = - 20 m$
$⟹ \frac{1}{v} + \frac{1}{- 60} = \frac{1}{- 20}$
$v = - 30 m$
Negative sign represents that object and image are on the same side of mirror.

Question 27.

If a light ray from air enters three different media in such a way that the refracted angles are $45^{0}$ , $30^{0}$ and $60^{0}$ respectively. Identify the right statement with respect to the optical density of these media.

Optical density is highest for medium 3.
Optical density is highest for medium 2.
Optical density is least for medium 3.
Optical density is least for medium 2.

Discuss Question

Answer: Option B. -> Optical density is highest for medium 2.
:
B
Angle of incidence

i

is same for all three media. Angle of refraction for the different media are:

r_{m 1} = 45^{\circ}

r_{m 2} = 30^{\circ}

r_{m 3} = 60^{\circ}

Deviation will be highest for medium with highest optical density. Clearly, the deviation is highest

(30^{\circ})

for Medium 2. Hence, Medium 2 has highest optical density.

Question 28.

Choose which of the following options are correct.

Discuss Question

Answer: Option A. ->

:
A and B
Convex lens has outward curved and concave lens has inward curved surfaces as shown in figure:

Choose Which Of The Following Options Are Correct.

A concave lens is thinner at its centre than at its edges, while a convex lens is thicker at it's centre than at it's edges.

Question 29.

For a light ray passing from air to water, according to Snell's law:
( $n$ represents refractive index of the medium, $θ_{a i r}$ is the angle of incidence and $θ_{w a t e r}$ is the angle of refraction)

$n_{a i r}$ $S i n θ_{a i r}$ = $n_{w a t e r}$ $S i n θ_{w a t e r}$
$n_{a i r}$ $S i n θ_{w a t e r}$ = $n_{w a t e r}$ $S i n θ_{a i r}$
$n_{a i r}$ $S i n θ_{a i r}$ $n_{w a t e r}$ $S i n θ_{w a t e r}$ = 1
$n_{a i r}$ $n_{w a t e r}$ = $S i n θ_{w a t e r}$ $S i n θ_{a i r}$

Discuss Question

Answer: Option A. ->

n_{a i r}

S i n θ_{a i r}

n_{w a t e r}

S i n θ_{w a t e r}

:
A
According to Snell's law, for a light ray travelling from one medium to another, the ratio of sines of angle of incidence to the angle of refraction is a constant. This constant is the refractive index of medium 2 with respect to medium one.
Here,

n_{a i r}

s i n θ_{a i r}

n_{w a t e r} s i n θ_{w a t e r}

\Rightarrow \frac{S i n θ_{a i r}}{S i n θ_{w a t e r}} = \frac{n_{w a t e r}}{n_{a i r}}

correctly represents Snell's law.

Question 30.

Two thin lenses of focal lengths $f_{1}$ and $f_{2}$ are placed in contact with each other. The focal length of the combination will be given by:

$\frac{f_{1} f_{2}}{f_{1} - f_{2}}$
$\sqrt{f_{1} f_{2}}$
$\frac{f_{1} f_{2}}{f_{1} + f_{2}}$
$\frac{f_{1} + f_{2}}{2}$

Discuss Question

Answer: Option C. ->

\frac{f_{1} f_{2}}{f_{1} + f_{2}}

:
C
The power of a lens is defined as the reciprocal of its focal length (in metres). When there is a combination of two lenses in contact, the effective power of the combination is the sum of the individual powers.
Total Power,