Physics Test 204

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Physics Test 204

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Weekly Quiz Competition

Question 1
4 / -1

What is the angular spread of the first minimum and the central maximum when diffraction is observed with 540 nm light through a slit of width 1mm?

A
540 x 10^-6 rad

B
1080 x 10^-6 rad

C
270 x 10^-6 rad

D
810 x 10^-6 rad

Solution

The First minimum is located at the place where the central maximum ends. And the first minimum spreads further upto half the fringe width.
Question 2
4 / -1

Two identical point masses P and Q, suspended from two separate massless springs of spring constant k₁ and k₂, respectively, oscillate vertically. If their maximum speeds are the same, the ratio (A_Q/A_P) of the amplitude A_Q of mass Q to the amplitude A_P of mass P is:

A

B

C

D

Solution
Question 3
4 / -1

Two waves are represented by the equations y₁ = a sin (ωt + kx + 0.57) m and y₂ = a cos (ωt + kx) m, where x is in meter and t in sec. The phase difference between them is

A
1.0 radian

B
1.25 radian

C
1.57 radian

D
0.57 radian

Solution
Question 4
4 / -1

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is x / 2 times its original time period. Then the value of x is:

A
√3

B
√2

C
√1

D
4

Solution
Question 5
4 / -1

A second pendulum is mounted in a space shuttle. Its period of oscillations will decrease when rocket is:

A
moving in geostationary orbit

B
ascending up with uniform acceleration

C
descending down with uniform acceleration

D
moving up with uniform velocity

Solution
Question 6
4 / -1

Fundamental note in open pipe (v₁= ν/2L) has _________ the frequency of the fundamental note in closed organ pipe (v₂= ν/4L).

A
Twice

B
Half

C
Four times

D
Thrice

Solution

Let L be a length of the pipe,

The fundamental frequency of closed pipe is

v₂=ν/4L .....(i)

where ν is the speed of sound in air.

Fundamental frequency of open pipe of same length is

v₁=ν/2L .....(ii)

After dividing v₁ with v₂,

v₁/v₂= ν/2L/ ν/4L

v₁=2v₂
Question 7
4 / -1

A lens of power + 2.0 D is placed in contact with another lens of power – 1.0 D. The combination will behave like

A
a converging lens of focal length 50 cm

B
a diverging lens of focal length 100 cm

C
a diverging lens of focal length 50 cm

D
a converging lens of focal length 100cm

Solution

P=P1+P2=+2−1=+1 dioptre, lens behaves as convergent

F=1/P=1/1=1m=100cm
Question 8
4 / -1

If the Young’s apparatus is immersed in water, the effect on fringe width will

A
remain same

B
decrease

C
increase

D
Cant say because the experiment cannot be carried in any other medium except air

Solution

When Young's double-slit set up for interference is shifted from air to within water then the fringe width decreases because the refractive index of water is more than that of the air.

Originally the fringe width is given by:

β1=λD/2d

The new fringe width within water will be given by

β2= λD/2nd

So, β2= β1/n

Here, n is the refractive index of medium.
Question 9
4 / -1

An unpolarised beam of intensity Io is incident on a polarizer and analyser placed in contact. The angle between the transmission axes of the polarizer and the analyser is θ. What is the intensity of light emerging out of the analyser?

A
(I₀ cos 2θ)/2

B
I₀ cos2θ

C
I₀ cosθ

D
I₀

Solution

Suppose the angle between the transmission axes of the analyser and the polarizer is θ. The completely plane polarized light form the polarizer is incident on the analyser. If E₀ is the amplitude of the electric vector transmitted by the polarizer, then intensity I₀ of the light incident on the analyser is

I ∞ E₀²

The electric field vector E₀ can be resolved into two rectangular components i.e E₀ cosθ and E₀ sinθ. The analyzer will transmit only the component ( i.e E₀ cosθ ) which is parallel to its transmission axis. However, the component E₀sinθ will be absorbed by the analyser. Therefore, the intensity I of light transmitted by the analyzer is,

I ∞ ( E₀ x cosθ )²

I / I₀ = ( E₀ x cosθ )² / E₀² = cos2θ

I = I₀ x cos²θ

when light passes from polarizer it's intensity becomes half and when passed through analyser it becomes,

I = I₀ x cos²θ/2
Question 10
4 / -1

What is the relation between critical angle and refractive index?

A
μ = cosC

B
μ = 1/cosC

C
μ = 1/sinC

D
μ = sinC

Solution

In Optics, the angle of incidence to which the angle of refraction is 90° is called the critical angle. The ratio of velocities of a light ray in the air to the given medium is a refractive index. Thus, the relation between the critical angle and refractive index can be established as the Critical angle is inversely proportional to the refractive index.

Critical Angle and Refractive Index

The relationship between critical angle and refractive index can be mathematically written as –

SinC=1/μ_a^b

μ_a^b =1/SinC

Where,

C is the critical angle.

μ is the refractive index of the medium.

a and b represent two mediums in which light ray travels.