Question

AIPMT 2015
In a double slit experiment, the two slits are
1 mm apart and the screen is placed 1 m away.
A monochromatic light of wavelength 500 nm is
used. What will be the width of each slit for obtaining
ten maxima of double slit within the central maxima
of single slit pattern ?
(1) 0.1 mm
(2) 0.5 mm
(3) 0.02 mm
(4) 0.2 mm​

Answer 1

Answer:

0.2mm

Explanation:

d=10^-3m

D=1m

lambda=500×10^-9

The width of central maxima in a single slit diffraction=2lamdaD/a

fringe width in double slit pattern β=lambdaD/d

10β=2lambdaD/a

10lambdaD/d=2lambdaD/a

a=d/5mm

a=10^-3/5=0.2mm

THANK YOU

Answer 2

Question:

In a double slit experiment, the two slits are 1 mm apart and the screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern ?

1. 0.1 mm
2. 0.5 mm
3. 0.02 mm
4. 0.2 mm

Answer:

(4) 0.2 mm

Solution:

$d = {10}{-3m}$

$D = 1m$

$λ = 500 \times {10}{-9} m$

The width of central maxima in a single slit diffraction: $\frac{a}{2λD}$

Fringe width in double slit pattern:  $β = \frac{d}{λD}$

Given,

$10β = \frac{a}{2λD}$

$\implies 10 \frac{d}{λD} = \frac{a}{2λD}$

$\implies a = \frac{5}{dmm}$

$\huge \boxed {\implies 0.2mm}$