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

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…..

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

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 ….

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

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

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

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° ….

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)

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

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

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

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

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

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

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