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Correct option is (A) 0.275 mm

Correct option is (D) 4 : 1.

Correct option is (D) 3 W.

Correct option is (D) 21 π rad

The phenomenon of bending of light waves around the edges (or corners) of the obstacles and entering into the expected geometrical shadow of the obstacle is called diffraction of light. 

Example: Colours observed when a CD (Compact Disc) is viewed is due to diffraction of light. 

(a), (b), (d)

(a) S₁ and S₂ have the same intensities. 

(b) S₁ and S₂ have a constant phase difference. 

(d) S₁ and S₂ have the same wavelength.

(A) 0.149 × 10⁷

n = tan i_B 

=> Brewster’s angle for glass: 

i_B = tan^–1 (n) = tan–1 (1.5) 

= 56° 19’.  

If the angle θ between the axes of polarization of the two polarizers, is 0°, i.e., the polarization axes of the two polarizers are parallel, the intensity of emergent light is maximum. If θ = 90°, i.e., the polarization axes of the two polarizers are perpendicular, no light emerges from the second polarizer, thus the intensity of emergent light is minimum. 

[Note: θ = 0° and 90° are known as parallel and cross settings of the two polarizers.]

Correct option is (D) diffraction.

Data : Refractive index, n = 1.5 

The reflected and refracted rays will be perpendicular to each other, when the angle of incidence = the polarizing angle θ_B,

tan θ_B = n = 1.5 

∴ θ_B = tan^-1 (1.5) = 56°19′

Correct option is (C) 60°

(A) increases with increase in wavelength

(a) The values shown by Neelam are : 

(i) high degree of general awareness, 

(ii) ability to convince someone, 

(iii) thinking skills, and 

(iv) concern for her friend. 

(b) The size of the obstacle should be comparable to the wavelength of the light wave in order to obtain an observable diffraction pattern. The size of the wall is 7 m, which is comparable enough with sound wave but not with the light wave. So, the two students cannot see each other but can talk to each other

(i) Reflection and refraction arise through interaction of incident light with atomic constituents of matter which vibrate with the same frequency as that of  the incident light. Hence frequency remains unchanged.

(ii) No. [Energy carried by a wave depends on the amplitude of the wave, not on the speed of wave propagation].

(iii) For a given frequency, intensity of light in the photon picture is determined by the number of photon incident normally on crossing a unit area per unit time.

Correct option is (B) 0.05 cm

(B) 2.5 × 10^-3 rad 

aⁿm = \(\cfrac{v_a}{v_m}\)

= \(\cfrac{v\lambda_a}{v\lambda_m}\) = \(\cfrac{5000}{4000}\)

= 1.25

is the refractive index of the medium with respect to air.

Data : c = 3 × 10⁸ m/s

(a), (b)

(a) the size decreases.
(b) the intensity increases.

Data : λ_a = 6400 Å, 

s_g = thickness of the glass slab = 5 cm, 

λ_g = 4000 Å

The speeds of light in glass and air are, respectively, v_g = λ_gv and v_a = λ_av where the frequency of the light v remains unchanged with the change of medium. 

The time taken to travel through the glass slab,

t_g = \(\cfrac{S_g}{v_g}\) = \(\cfrac{S_g}{λ_gv}\)

The time taken to travel through air,

 t_a = \(\cfrac{S_a}{v_a}\) = \(\cfrac{d}{λ_av}\)

where s_a = d is the distance of the glass surface from the source.

Since t_a = t_g

\(\cfrac{d}{λ_av}\) = \(\cfrac{S_g}{λ_gv}\)

∴ d = \(\cfrac{λ_a}{λ_g}\) S_gS_g = \(\frac{6400}{4000}\) x 5 = 1.6 × 5 = 8 cm

(i) Spherical wavefront 

(ii) Plane wavefront. 

(a) Dependence of the speed of light in a vacuum: The speed of light in vacuum is a universal constant. It is independent of all the factors listed above. It may be pointed out that Newton ion relativity principle does not apply to the speed of light. The fact that the speed of light is independent of the relative motion between the source and the observer is one of the postulates of Einstein’s special theory of relativity.

(b) Dependence of the speed of light in a medium:

(i) It does not depend on the direction of propagation, provided the medium is isotropic.

(ii) It does not depend on the relative motion between the source and the medium. However, it depends on relative emotion between, the source and the observer

(iii) It depends on the wavelength

(iv) It does not depend on the intensity.

Correct option is (B) spherical

Doppler effect for sound waves is asymmetric i.e. for the same relative motion in the cases (i) and (ii), the apparent change in the frequency is not the same. It is because, sound waves require a medium for their propagation and the motion of the observer relative to the medium is different in two cases.

Doppler effect for light waves is symmetric i.e. for the same relative motion in the case
(i) and (ii), the apparent change in the frequency is same. For light waves in vacuum, there is no difference in the two cases. It is because, the relative motion between the source and the observer is not affected, whether the source is in motion or the observer is in motion. However, for the light waves in medium, the situations in  the two cases are different as in the case of Doppler effect for sound waves. Likewise the formulas for Doppler effect in the two cases will be different.

The width of central maximum \(\frac{2D\lambda}{a}\)

If the width of the slit is made double of the original, the width of the central diffraction band will reduce to half. The intensity of the central diffraction band coil become four times.

Correct option is (A) spherical

Correct option is (B) v ∆ t

X = 5000 A = 5 x 10^-7 m
on reflection, no change occurs in wavelength, frequency and speed of the incident light.
∴ Wavelength of reflected light = 5000 Å.
Taking C = 3 x 10⁸ ms^-1. The frequency of reflected light is given by

V = \(\frac{C}{\lambda}\) = \(\frac{3\times 10^8}{5 \times 10^7}\) = 6 x 10¹⁴ Hz

The reflected ray is normal to the incident ray, when the angle of incidence is equal to 45°.

(B) considerably larger than λ

(C) π, 3π, 5π, …

Correct option is (A) zero

µ = 1.5

According to Brewster’s law,

µ = tan p

tan p = 1.5

⇒ p = 56.31º

n = tan i_B 

=> Brewster’s angle for glass: 

i_B = tan^–1 (n) 

= tan^–1 (1.5) = 56° 19’.

Given λ = 5000 Å = 5 × 10^-7 m

The wavelength and frequency of reflected light remains same.

∴ Wavelength of reflected light,

λ = 5000 Å.

Frequency of reflected light,

v = \(\frac{c}{λ}\)  = \(\frac{3\times10^8}{5\times10^{-7}}\)

= 6 × 10¹⁴ HZ 

The reflected ray is normal to incident if angle of incidence i = 45°.

1. It is spherical wavefront,

2. It is plane wavefront.

3. Plane wavefront (a small area on the surface of a large sphere is nearly planar).

(D) diverging or converging.

Given µ = 1.5, θ = ?

Since µ = tan θ

∴ tan θ = 1.5

or θ = tan^-1 1.5 = 56.3°.

(i) θ = 42°

(ii) P = 57°

(iii) Straight line

(iv) Planks constant

(v) µ_r = 1.2

(vi) Paramagnetic.

1. Interference

2.  β₁ = \(\frac{λ_1D}{d},β_2=\frac{λ_2D}{d}\)  \(\frac{β_1}{β_2}=\frac{λ_1}{λ_2}\)

λ₁ = 5461 A° = 5461 × 10^-10 m
β₁ = 0.15mm = 0.15 × 10^-3m
β₂ = 0.12mm = 0.12 × 10^-3m.

∴ λ₂ = \(\frac{β_2\timesλ_1}{f_1}\)

= \(\frac{546\times10^{-10}\times0.12\times10^{-3}}{0.15\times10^{-3}}\) = 4368A⁰

4. Wave length of blue is less than that of red. Hence β decreases.

1. β = \(\frac{λD}{d}\)

2. The condition for dark fringe is P₂R – P₁R = (2n -1)\(\frac{λ}{2}\) where n = 1,2, 3, etc.

3. No. They are not coherent sources.

4.  β =  \(\frac{λD}{d}\) = \(\frac{1\times5\times10^{-7}}{10^{-3}}\) = 5 x 10^-4 = 0.5 mm

1. Spherical wave front

2. Cylindrical wavefront

3. Diverging wavefront

4. Plane wavefront

Correct option is: (A) The interference pattern will remain unchanged

1. Width of bands decreases from the centre of the bands

2. The pattern expands or bandwidth increases

3. X_n = \(\frac{nλD}{d}\)

d = 0.03 cm = 0.03 × 10^-2 m

D = 1.5 m, n = 4, x_n = 1 cm = 1 × 10^-2 m

λ = \(\frac{X_nd}{nD}\) = \(\frac{10^{-2}\times0.03\times10^{-2}}{4\times1.5}\) = 500 nm

(c) 141 nm

Explanation: Here, 2µt = \(\frac{λ}{2}\)

∴ t_min = \(\frac{λ}{4\mu}=\frac{750}{4\times1.33}\) = 141 nm

The nature of the wavefronts emerging from the final image is Spherical.

Yes, Huygen’s principle valid for longitudunal sound waves

Principle of superposition of waves : The displacement at a point due to the combined effect of a number of waves arriving simultaneously at the point is the vector sum of the displacements due to the individual waves arriving at the point.

Explanation : This is a general principle of linear systems applied to wave phenomena. When two or more wave trains arrive at a point simultaneously, they interpenetrate without disturbing each other. The resultant displacement at that point is the vector sum of the displacements due to the individual waves arriving at the point. The amplitude and phase angle of the resulting disturbance are functions of the individual amplitudes and phases.

Notes :

1. In the case of mechanical waves, e.g., sound, the displacement is that of a vibrating particle of the medium. 

2. When we consider superposition of electromagnetic waves, light in this chapter, the term displacement refers to the electric field component of the electromagnetic wave.

(i) Spherical converging wavefront; 

(ii) Plane wavefront. 

(a) Plane wavefront; 

(b) Plane wavefront.