Question

In Young's double slit experiment, red light of wavelength
6000A is used and at point P on the screen, nth bright
band is obtained. Keeping the set-up of the experiment
same, now green light of wavelength 5000A is used and at
point P on the screen, (n+1) th bright band is obtained.
Find the magnitude of n.
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Answer 1

Answer:

n=5

Explanation:

We have,

nλ = xd/D  

Where n = number of fringe from central fringe.

x= distance of nth fringe from nth fringe i.e distance from central fringe to point P here.

λ = wavelength of wave.

d= distance between slits

D= Distance between screen and slits.

Now, for 6000A radiation,

$\\\tt n_{red} \lambda _{red} = xd/D\\\\\tt n_{red} (6 \times 10^{-7} m ) = xd/D$       -----------(i)

Since the distance to P from central fringe is constant here x will be same for both cases,

Now for wave 5000A,

$\\\tt n_{green} (5 \times 10^{-7} m ) = xd/D$      --------------(ii)

Dividing  (i) by (ii)

$\\\tt \dfrac{ n_{red} (6 \times 10^{-7} m ) }{n_{green}(5 \times 10^{-7} m ) } =\dfrac{x}x} =1\\\implies \dfrac{ (6 \times 10^{-7} m ) }{ (5 \times 10^{-7} m ) } = \dfrac{n+1}{n}$

Solving this n = 5.