Wave Optics

A man wants to distinguish between two pillars located at a distance of 11 km. What should be the minimum distance between the pillars?

by

Q: A man wants to distinguish between two pillars located at a distance of 11 km. What should be the minimum distance between the pillars ?

Sol: As the limit of resolution of human eye is

$\displaystyle \theta = 1′ = (\frac{1}{60})^o $

$\displaystyle \theta = \frac{\pi}{60 \times 180} $

If x is minimum distance between the pillars,

and d = 11 km = 11 × 103 m ,

then from $\displaystyle \theta = \frac{x}{d} $

$\displaystyle x = d \times \theta $

$\displaystyle x = 11 \times 10^3 \times \frac{\pi}{60 \times 180} $

x = 3.2 m

Two beams, A and B of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid…..

by

Q: Two beams, A and B of plane polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when the beam A has maximum intensity. (and beam B has zero intensity), a rotation of polaroid through 30° makes the two beams appear equally bright. If the initial intensities of the two beams are IA and IB respectively, then IA/IB equals.

(a) 1

(b) 1/3

(c) 3

(d) 3/2

Click to See Answer :
Ans: (b)

Sol: As IA and IB are initial intensities, therefore, on rotation of polaroid through 30°,

IA‘ = IA cos2⁡30°

IB‘ = IB cos2⁡60°

As IA‘ = IB‘

IA cos2⁡30° = IB cos2⁡60°

$\displaystyle \frac{I_A}{I_B} = \frac{cos^2 60^o}{cos^2 30^o} $

$\displaystyle \frac{I_A}{I_B} = \frac{1/4}{3/4} $

$\displaystyle \frac{I_A}{I_B} = \frac{1}{3} $

 

A beam of light λ = 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern ….

by

Q: A beam of light λ = 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central bright fringe is

(a) 1.2 cm

(b) 1.2 mm

(c) 2.4 cm

(d) 2.4 mm

Click to See Answer :
Ans: (d)
Sol: Here, λ = 600 nm = 600 × 10-9 m
a =1 mm = 10-3 m , D = 2 m
Distance between first dark fringes on either side of central bright fringe is width of central maximum, which is

$\displaystyle 2 x = \frac{2 \lambda D}{a} $

$\displaystyle 2 x = \frac{2 \times 6 \times 10^{-7}\times 2}{10^{-3}} $

$\displaystyle 2 x = 2.4 \times 10^{-3}m = 2.4 mm $

 

A beam of unpolarized light of intensity I_0 is passed through a polaroid A and then through another polaroid

by

Q: A beam of unpolarized light of intensity I0 is passed through a polaroid A and then through another polaroid B which is oriented so that it principal plane makes an angle of 45° relative to that of A . The intensity of emergent light is

(a) I0

(b) I0/2

(c) I0/4

(d) I0/8

Click to See Answer :
Ans: (c)
Sol: Here, polaroid A polarizes light.

Intensity of polarized light from A= I0/2

According to law of Malus ,

Intensity of light emerging from B,

$\displaystyle I_B = \frac{I_0}{2}cos^2 45^o $

$\displaystyle I_B = \frac{I_0}{2}(\frac{1}{\sqrt{2}})^2 $

$\displaystyle I_B = \frac{I_0}{4} $

 

In a diffraction pattern due to a single slit of width a , the first minimum is observed at an angle of 30° ….

by

Q: In a diffraction pattern due to a single slit of width a , the first minimum is observed at an angle of 30° . When light of wavelength 5000 A ̇ is incident on the slit. The first secondary maximum is observed at an angle of

(a) sin-1(2/3)

(b) sin-1(1/2)

(c) sin-1(3/4)

(d) sin-1(1/4)

Click to See Answer :
Ans: (c)

Sol:  For first minimum,

a sin⁡θ = 1 λ , ( since n = 1 )

or , a sin⁡30° = λ

or , a/2 = λ

or , a = 2 λ

For first secondary maximum

$\displaystyle a sin\theta’ = (2n+1)\frac{\lambda}{2} $

$\displaystyle a sin\theta’ = \frac{3 \lambda}{2} $

$\displaystyle sin\theta’ = \frac{3 \lambda}{2 a} $

$\displaystyle sin\theta’ = \frac{3 \lambda}{2 \times 2 \lambda} $

$\displaystyle sin\theta’ = \frac{3}{4} $

$\displaystyle \theta’ = sin^{-1}(\frac{3}{4} )$