Ray Optics and Optical Instruments Test

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Ray Optics and Optical Instruments Test - 4

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Weekly Quiz Competition

Question 1
1 / -0.25

The magnifying power of the telescope can be increased by

A
increasing the focal length of eyepiece

B
fitting eye-piece of low power

C
increasing the distance of an object

D
fitting eye-piece of high power

Solution

Magnifying power of a telescope is f₀ /f_e , so as 1/f_e increases, magnifying power increases.
Hence the correct answer is option D.
Question 2
1 / -0.25

A lamp and a screen are set up 100 cm apart and a convex lens is placed between them. The two positions of the lens forming real images on the screen are 40 cm apart. What is the focal length of the lens ?

A
21 cm

B
15 cm

C
18 cm

D
12 cm

Solution

L₁ L₂ =40cm and O I=100cm
Therefore, x+40+x=100 or x=30cm
For lens at L1, we have
U=-30cm and v=+70
Thus,
(1/v)-(1/u)=1/f
Or, 1/f=(1/70) - (1/-30)
Or, f=+21cm
Question 3
1 / -0.25

The magnifying power of telescope is high if

A
the objective has a short focal length and the eye-piece has a long focal length

B
both objective and eye-piece have short focal lengths

C
the objective has a long focal length and the eye-piece has a short focal length

D
both objective and eye-piece have long focal length
Question 4
1 / -0.25

Refraction is

A
Reversal of direction of propagation of light

B
The change of direction of propagation of light

C
None of the above

D
Change of direction of propagation of light at the interface of two media as light passes from one medium to another
Question 5
1 / -0.25

Accommodation of the human eye is

A
Ability to see an erect image

B
Ability to change the focal length

C
Ability to change the intensity of incoming light

D
Ability to see close by objects

Solution

The ability of the eye lens to adjust its focal length is called power of accommodation. This is done by the ciliary muscles by changing the focal length of the eye lens.
Question 6
1 / -0.25

The largest telescope in the world has a reflector with an aperture of 200 inches in order to achieve

A
low dispersive power

B
high accommodation power

C
least spherical aberration

D
high resolving power
Question 7
1 / -0.25

A convex lens forms a real image of an object on a screen; the magnification of the image being 3/2. The object and the screen are kept fixed and the lens is moved through a distance of 16 cm when a sharp image is again formed on the screen; the magnification now being 2/3. What is the focal length of the lens?

A
19.6 cm

B
19.2 cm

C
20 cm

D
20.4 cm

Solution

3/2=v/u …...1
2/3=(v-16)/(u+16)......2
from these equation
we get v=48, u=32
so now f=(vxu)/(v+u)=96/5=19.2
The correct answer is option B.
Question 8
1 / -0.25

A plano-convex lens, when silvered on the plane side, behaves like a concave mirror of focal length 30 cm. When it is silvered on the convex side, it behaves like a concave mirror of focal length 10 cm. The refractive index of the material of the lens is

A
2.0

B
2.5

C
1.5

D
3.0

Solution

Given: A plano-convex lens when silvered on the plane side behaves like a concave mirror of focal length 60cm. However, when silvered on the convex side, it behaves like a concave mirror of focal length 20cm.
To find the refractive index of the lens
We know,
1/F =(2/f )+(1/f_m )............(i), where
When plane surface is silvered it becomes concave mirror of focal length 60cm,
and f_m =∞
So the eqn(i) becomes,
1/60 =(2/f )+(1/∞)
⟹f=120cm
When convex side is silvered it becomes concave mirror of focal length 10cm,
and f_m =R/2
So the eqn(i) becomes,
1/20 =2/f +2/R
⟹2/R=(1/20 )−(2/f)
⟹2/R =(1/20)−(2/120)
⟹2/R=(6 −2 /120)
⟹R=60cm
Now using the focal length formula of the plano-convex lens, we get
1/f =(μ−1)×1/R
⟹1/120=(μ−1)×1/60
⟹μ−1=(60/120)
⟹μ=(1/2)+1
⟹μ=1.5
is the refractive index of the lens.
Question 9
1 / -0.25

For a rectangular slab, refraction takes place at

A
Two interfaces

B
Three interfaces

C
Four interfaces

D
One interfaces

Solution

The refraction takes place at both the air-glass interface and glass-air interface of a rectangular glass slab. When the light ray incident on the air-glass interface (DC) obliquely, it bends towards the normal.
Question 10
1 / -0.25

Farsighted or hypermetropic eye can be corrected by

A
diverging lens

B
converging (positive) lens

C
cylindrical lens

D
Plano-concave lens

Solution

Hypermetropia is corrected using converging lens. In this defect the person is able to see far objects clearly but difficulty with near vision. The image is focused behind the retina rather than upon it. This occurs when eyeball is too short or the power of lens is too weak. By using a convex lens the defect is corrected.

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Re-Attempt Weekly Quiz Competition

Ray Optics and Optical Instruments Test - 4

Result

Ray Optics and Optical Instruments Test - 4

Score

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Rank

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SHARING IS CARING

Question 1

increasing the focal length of eyepiece

fitting eye-piece of low power

increasing the distance of an object

fitting eye-piece of high power

Solution

Question 2

21 cm

15 cm

18 cm

12 cm

Solution

Question 3

the objective has a short focal length and the eye-piece has a long focal length

both objective and eye-piece have short focal lengths

the objective has a long focal length and the eye-piece has a short focal length

both objective and eye-piece have long focal length

Question 4

Reversal of direction of propagation of light

The change of direction of propagation of light

None of the above

Change of direction of propagation of light at the interface of two media as light passes from one medium to another

Question 5

Ability to see an erect image

Ability to change the focal length

Ability to change the intensity of incoming light

Ability to see close by objects

Solution

Question 6

low dispersive power

high accommodation power

least spherical aberration

high resolving power

Question 7

19.6 cm

19.2 cm

20 cm

20.4 cm

Solution

Question 8

2.0

2.5

1.5

3.0

Solution

Question 9

Two interfaces

Three interfaces

Four interfaces

One interfaces

Solution

Question 10

diverging lens

converging (positive) lens

cylindrical lens

Plano-concave lens

Solution

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