Question

(4) Completly polarised and poorly intense
NEET-UG 2013
In Young's double slit experiment, the slits are 2mm
apart and are illuminated by photons of two
wavelengths 21 = 120008 and 12 = 10000Å. At
what minimum distance from the common central
bright fringe on the screen 2m from the slit will a
bright fringe from one interference pattern coincide
with a bright fringe from the other ?
(1) 3 mm (2) 8 mm (3) 6 mm (4) 4 mm
​

Answer 1

Answer: option c

given they both coincide

hence   n1 w1= n2 w2                  n= no of fringe,  w= wavelength

by applying we get n1/n2= 5/6

take n1=5 apply in equation

distance= n1* wi* D/d             d= distance b/w slits

                                                 d- distance b/w screen and the mirror

Explanation:

Answer 2

Answer:

(3) 6mm

Explanation:

I am writing lambda using this sign " ¥ "

Let n1 bright Fringe of ¥1 coincides with n2 bright fringe of ¥2.then

n1 ¥1 D/d=n2 ¥2 D/d

or, n1 ¥1 =n2 ¥2

or, n1/n2=¥1/¥2=10000/12000=5/6

let, x be given distance.

so, x=n1 ¥1 D/d

here,

n1=5,D=2m,d=2mm=2×10^-3m

¥1=12000 angstrom = 12000×10^-10m=12×10^-7m

x= (5×12×10^-7×2)/2×10^-3=6×10^-3m=6mm

Hope it will help you.

please mark me as brainliest answer.