Q: A parallel beam of light strikes a piece of transparent glass having cross section as shown in the figure below . Correct shape of the emergent wavefront will be

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Q: A parallel beam of light strikes a piece of transparent glass having cross section as shown in the figure below . Correct shape of the emergent wavefront will be

Ans: (a)

Sol:

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 × 10^{3} 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

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 I_{A} and I_{B} respectively, then I_{A}/I_{B} equals.

(a) 1

(b) 1/3

(c) 3

(d) 3/2

Ans: (b)

Sol: As I_{A} and I_{B} are initial intensities, therefore, on rotation of polaroid through 30°,

I_{A}‘ = I_{A} cos^{2}30°

I_{B}‘ = I_{B} cos^{2}60°

As I_{A}‘ = I_{B}‘

I_{A} cos^{2}30° = I_{B} cos^{2}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} $

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 $