Wave Optics Test

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Wave Optics Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

Choose the Correct answer from alternative given.
Wavefront is the locus of all points, where the particles of the medium vibrate with the same

A
phase

B
amplitude

C
frequency

D
period

Solution

Wavefront is the locus of all points, where the particles of the medium vibrate with the same phase.
Question 2
1 / -0

The spectral line for a given element of light received from a distant star is shifted towards longer wavelength side by 0.025%. The velocity of star in the line of light is then

A
$$7.5 \times 10^4m s^{-1}$$

B
$$-7.5 \times 10^4m s^{-1}$$

C
$$3.7 \times 10^4m s^{-1}$$

D
$$-3.7 \times 10^4m s^{-1}$$

Solution

Given: The shift occurred in the wavelength of the light received from the star is $$0.025\%$$

To find: The velocity of the star in the line of light.

Since the wavelength of the light received is increased, it means that the star is traveling towards the observer.
Velocity of star in line of light is
$$v=\dfrac{-\Delta\lambda}{\lambda}\times c$$

The changed in wavelength is $$0.025\%$$ of the actual wavelength. So, $$\dfrac{\Delta\lambda}{\lambda}$$ will be $$\dfrac{0.025}{100}$$.

$$=-{\dfrac {0.025}{100}}\times 3\times 10^8=-7.5\times 10^4ms^{-1} $$.
Question 3
1 / -0

The human eye has an approximate angular resolution of $$\phi = 5.8 \times 10^{-4}$$rad and typical photoprinter prints a minimum of 300 dpi (dots per inch, 1 inch = 2.54 cm). At what minimal distance z should a printed page be held so that one does not see the individual dots?

A
14.5 cm

B
20.5 cm

C
29.5 cm

D
28 cm

Solution

Here, angular resolution of human eye,
$$\phi \, = \, 5.8 \, \times \, 10^{-4} \, red$$
The linear distance between two successive dots in a typical photo printer is $$l \, = \, \dfrac{2.54}{300} \, cm \, = \, 0.84 \, \times \, 10^{-2} \, cm.$$
At a distance of z cm, the gap distance l will subtend an angle
$$\phi \, = \, \dfrac{l}{z} \, \therefore \, z \, = \, \dfrac{l}{\phi} \, = \, \dfrac{0.84 \, \times \, 10^{-2} \, cm}{5.8 \, \times \, 10^{-4}} \, = \, 14.5 \, cm$$
Question 4
1 / -0

What is the minimum thickness of a soap film needed for constructive interference in reflected light, if the light incident on the film is $$\text{750 nm}$$? Assume that the refractive index for the film is $$\mu \, = \, 1.33.$$

A
$$\text{282 nm}$$

B
$$\text{70.5 nm}$$

C
$$\text{141 nm}$$

D
$$\text{387 nm}$$

Solution

Given: The wavelength of the light incident on the soap film is $$750\ nm$$.
The refractive index of the film is $$1.33$$.

To find: The minimum thickness of the soap film required for constructive interference.

The thickness of the film for the constructive interference is given by:
$$2t=\left(m+\dfrac12\right)\lambda'$$

The new wavelength in the medium is given as:
$$\lambda'=\dfrac{\lambda}{\mu}$$

So, the minimum thickness of the film $$'t'$$ will be at $$m=0$$ :
$$2t_{min}=\left(0+\dfrac12\right)\dfrac{750\times10^{-9}}{1.33}$$

$$t_{min}=\dfrac{750\times10^{-9}}{4\times1.33}$$

$$=1.41\times10^{-7}\ or\ 141\ nm$$
Question 5
1 / -0

Wavefront is the locus of all points, where the particles of the medium vibrate with the same.

A
phase

B
amplitude

C
frequency

D
period

Solution

Wavefront is the locus of all points, where the particles of the medium vibrate with the same phase
Question 6
1 / -0

Transverse nature of light was confirmed by the phenomenon of the

A
refraction of light

B
diffraction of light

C
dispersion of light

D
polarization of light.

Solution

The phenomenon of polarization confirms that light is a transverse wave because for polarization, the light should have different components oscillating in the different planes and a transverse wave has the oscillations perpendicular to the direction of propagation of the wave.
Question 7
1 / -0

The diameter of the pupil of human eye is about 2 mm. Human eye is most sensitive to the wavelength of 555 nm. The limit of resolution of human eye is:

A
1.2 min

B
2.4 min

C
0.6 min

D
0.3 min

Solution

Limit of resolution
$$d\theta = \dfrac{1.22\lambda}{D} = \dfrac{1.22\times555\times10^{-9}}{2\times10^{-3}}$$ = 3.39 $$\times 10^{-4}$$ rad

$$d\theta = 3.39 \times 10^{-4} \times \left[\dfrac{180^{\circ}}{\pi}\right] $$

$$=0.0194^{\circ} = 0.0194 \times{60}\min = {1.2}min$$
Question 8
1 / -0

A slit of width a is illuminated by white light. The first minimum for red light $$(\lambda \, = \, 6500 \, \overset{0}{A})$$ will fall at $$\theta \, = \, 30$$ when a will be:

A
3200 $$\overset{0}{A}$$

B
$$6.5 \, \times \, 10^{-4} \, mm$$

C
1.3 micron

D
$$2.6 \, \times \, 10^{-4} \, cm$$

Solution

Given: The wavelength of the red light is $$6500\ A^o$$.
The angle at which the light falls is $$30^\circ$$.

To find: The width of the slit.

Path difference between the beams of the light is:
$$\Delta x =n \lambda$$

For first minimum, $$n=1$$
$$\Delta x=a \times sin \theta = 1 \lambda$$

$$a \, = \, \dfrac{\lambda}{sin\theta } \, = \, \dfrac{6.5 \times 10^{-7}}{sin 30^o}$$

$$ = 13 \times 10^{-7} \, = \, 1.3\, micron$$
Question 9
1 / -0

In Young's double slit eperiment two disturbances arriving at a point P have phase difference of $$\dfrac{\pi}{3}$$ . The intensity of this point expressed as a fraction of maximum intensity $$I_0$$ is then

A
$$\dfrac{3}{2}I_0$$

B
$$\dfrac{1}{2}I_0$$

C
$$\dfrac{4}{3}I_0$$

D
$$\dfrac{3}{4}I_0$$

Solution

The resultant intensity ,$$ I=I_o cos^2 \dfrac{\phi}{2}$$
Here, $$I_0$$ is the maximum intensity and $$\phi$$=$$\dfrac{\pi}{3}$$
$$\therefore I=I_0 cos^2$$
$$I=I_0 cos^2 \dfrac{\pi}{3 \times 2}=\dfrac{3}{4}I_0$$
Question 10
1 / -0

A parallel beam of sodium light of wavelength 5890 $$\overset{0}{A}$$ is incident on a thin glass plate of refractive index 1.5 such that the angle of refraction in the plate is $$60^o$$. The smallest thickness of the plate which will make it dark by reflection:

A
3926 $$\overset{0}{A}$$

B
4353 $$\overset{0}{A}$$

C
1396 $$\overset{0}{A}$$

D
1921 $$\overset{0}{A}$$

Solution

Let the thickness of the Glass slab is $$t$$

The condition for minimum thickness corresponding to a dark band is $$2\mu t \,cosr = \lambda$$

$$ \therefore t= \dfrac{\lambda}{2 \mu cos \,r} = \dfrac{5890 \times 10^{-10}}{2 \times 1.5 \times cos 60^o} $$

$$t=3926 \times 10^{-10} m = 3926 {\overset{o}A} $$

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Wave Optics Test - 44

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Wave Optics Test - 44

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SHARING IS CARING

Question 1

phase

amplitude

frequency

period

Solution

Question 2

$$7.5 \times 10^4m s^{-1}$$

$$-7.5 \times 10^4m s^{-1}$$

$$3.7 \times 10^4m s^{-1}$$

$$-3.7 \times 10^4m s^{-1}$$

Solution

Question 3

14.5 cm

20.5 cm

29.5 cm

28 cm

Solution

Question 4

$$\text{282 nm}$$

$$\text{70.5 nm}$$

$$\text{141 nm}$$

$$\text{387 nm}$$

Solution

Question 5

phase

amplitude

frequency

period

Solution

Question 6

refraction of light

diffraction of light

dispersion of light

polarization of light.

Solution

Question 7

1.2 min

2.4 min

0.6 min

0.3 min

Solution

Question 8

3200 $$\overset{0}{A}$$

$$6.5 \, \times \, 10^{-4} \, mm$$

1.3 micron

$$2.6 \, \times \, 10^{-4} \, cm$$

Solution

Question 9

$$\dfrac{3}{2}I_0$$

$$\dfrac{1}{2}I_0$$

$$\dfrac{4}{3}I_0$$

$$\dfrac{3}{4}I_0$$

Solution

Question 10

3926 $$\overset{0}{A}$$

4353 $$\overset{0}{A}$$

1396 $$\overset{0}{A}$$

1921 $$\overset{0}{A}$$

Solution

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