Question

light passes through narrow slits with a separation of 0.4mm.On a screen 1.6 away,the distance b/w 2 second order maximum 2.5mm. what is wavelength of light used

Answer 1

Answer:

ANSWER

Separation is given by

y=  

d

nλD

​  

 

where

d=3mm=3×10  

−3

 

D=2m

λ  

1

​  

=480nm=480×10  

−9

m

λ  

2

​  

=600nm=600×10  

−9

m

n  

1

​  

=n  

2

​  

=5

y  

2

​  

−y  

1

​  

=?

So,

y  

1

​  

=  

d

nλ  

1

​  

D

​  

 

y  

1

​  

=  

3×10  

−3

 

5×480×10  

−9

×2

​  

 

y  

1

​  

=1.6×10  

−3

m

Also,  

y  

2

​  

=  

d

nλ  

2

​  

D

​  

 

y  

2

​  

=  

3×10  

−3

 

5×600×10  

−9

×2

​  

 

y  

2

​  

=2×10  

−3

m

As y  

2

​  

>y  

1

​  

 

y  

2

​  

−y  

1

​  

=2×10  

−3

−1.6×10  

−3

 

             =4×10  

−

4m

Therefore the separation on the screen between the fifth order bright fringes of the two interference patterns is 4×10  

−4

m

Explanation: