Ray Optics and Optical Instruments Test

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Ray Optics and Optical Instruments Test - 12

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Question 1
1 / -0.25

An object is initially at a distance of 100 cm from a plane mirror. If the mirror approaches the object at a speed of 5 cm/s. Then after 6 s the distance between the object and its image will be

A
60 cm

B
140 cm

C
170 cm

D
150 cm

Solution

Given that,
The distance of object from plane mirror D=100cm
So, initial distance of image and object
=100 −(−100)
=200cm
Now, object approaches the mirror with the speed=v=5cm/s
So, distance travelled by object in 6 sec
D=5 ×6
D=30cm
At this instant,
Distance between mirror and object
D=100 −30
D=70cm
Now, the distance between image and object is
D=70 −(−70)
D=140cm
Hence, the distance between the object and its image is 140 cm.
Question 2
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Two mirrors are placed perpendicular to each other. A ray strikes the first mirror and after reflection from the first mirror it falls on the second mirror. The ray after reflection from second mirror will emerge

A
Perpendicular to the original ray

B
Parallel to the original ray

C
At 45⁰ to the original ray

D
At 60⁰ to the original ray

Solution

Let ray strikes the vertical mirror by angle of incidence θ. Then by joining the perpendiculars of two mirrors it is clear from the diagram that angle of incidence at the horizontal mirror is (90 −θ).

∴reflected ray from the horizontal mirror makes angle θwith the horizontal mirror also incident ray at vertical mirror makes angle θwith horizontal axis.
∴reflected ray is parallel to the original ray.
This is true for two perpendicular mirrors fixed at any positions.
Question 3
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A person is in a room whose ceiling and two adjacent walls are mirrors. How many images are formed?

A
5

B
6

C
7

D
8

Solution

The number of images formed when two mirrors are inclined to each other is given by :
n=(360/θ-1)
here θ= 90 °[since walls are perpendicular]
so, number of images=n=360/90-1
=4-1
=3
These 3 images are formed by a combination of two adjacent walls with the object itself acts as objects for the ceiling mirror. So totally 4 images are formed.
Hence total number of images formed are 4+3= 7
Question 4
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If an object is placed unsymmetrically between two plane mirrors, inclined at the angle of 60⁰ , then the total number of images formed is

A
5

B
4

C
2

D
infinite

Solution

The number of images between two plane mirror inclined at an angle θwhen the object is placed symmetrically between the mirrors is given as
n=(360o / θ) −1
Given,
θ=60o
∴n=(360^o /60)−1=6 −1=5
Question 5
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In image formation from spherical mirrors, only paraxial rays are considered because they

A
are easy to handle geometrically

B
contain most of the intensity of the incident light

C
form nearly a point image of a point source

D
show minimum dispersion effect.

Solution

Paraxial rays are ones which originate from a point source and make a small angle with the principal axis. After reflection, they give a point image, either real or virtual.
But if rays from point source that are far from the mirror axis, gives a blurred image, an effect called spherical aberration.
Question 6
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A concave mirror of radius of curvature 20 cm forms image of the sun. The diameter of the sun subtends an angle 1 °on the earth. Then the diameter of the image is (in cm)

A
2 π/9

B
π/9

C
20

D
π/18

Solution

Given: A concave mirror of radius of curvature 20cm forms image of the sun. The diameter of the sun subtends an angle 1o on the earth.

To find the diameter of the image

Solution:

As sun is more far from the earth,

Then we take u = infinite

Radius of curvature, R=−20cm

Focal length,

From the mirror formula

d = π/18
Question 7
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A convex mirror has a focal length f . A real object, placed at a distance f in front of it from the pole, produces an image at

A
2 f

B
f /2

C
f

D
¥

Solution

Here F=f
U=-f(-because it is present in front of a convex lens)
It is given that both focal length and object distance is f .
So according to lens formula 1/F=1/v+1/u
So 1/f=(1/v)+(-1/f)
=>1/f=(1/v)-(1/f)
So (1/f)+(1/f)=1/f
=>1+(1/f)=(1/v)
=>2/f=1/v
So from here we can say that v= f/2
Hence option B is correct.
Question 8
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A convex mirror has a focal length = 20 cm. A convergent beam tending to converge to a point 20 cm behind convex mirror on principal axis falls on it. The image if formed at

A
infinity

B
40 cm

C
20 cm

D
10 cm

Solution

The converging ray appears to converge on the focus of the convex mirror. Therefore after reflection they will travel parallel to the principal axis. So the image will be formed at infinity.

Hence,
option (A) is the correct answer.
Question 9
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An object is placed at a distance u from a concave mirror and its real image is received on a screen placed at a distance of v from the mirror. If f is the focal length of the mirror, then the graph between 1/v versus 1/u is

A

B

C

D

Solution

For real object and real Image,
(1/v)+(1/u)=(1/f)=>-(1/v)-(1/u)=-(1/f)
(1/v)+(1/u)=(1/f) =>1/v=-(1/u)+(1/f)
y=mx+c
The correct answer is option B.
Question 10
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An infinitely long rod lies along the axis of a concave mirror of focal length f. The near end of the rod is at a distance u >f from the mirror. Its image will have a length

A
f² /(u ^_ f)

B
uf/(u ^_ f)

C
f² /(u + f)

D
uf /(u + f)

Solution

At infinity, image of B will form at focus since it is at infinity. Image of A will be at A which can be calculated by mirror formula.
1/v +(1/−u)=(1/−f)
⟹v= fu /(f −u)=−( fu/(u −f))
Image length = v −f=(fu/(u −f))−(f)

= f² /(u −f)
(We take the absolute values of the distances to calculate the rod length)