Question

In a ydse bichromatic light of wavelengths 400 nm and 560 nm are used . The distance between the slits is 0.1 mm and the distance between the plane of the slits and the screen is 1 m . The minimum distance between two successive regions of complete darkness is 4 mm

Answer 1

Let nth minima of 400 nm wave coincides with mth minima of 560 nm wave.

Then,

(2n-1)(400/2) - (2m-1)(560/2)

2n-1 /2m-1  = 7/5 =14/10 ....

2n-1= 7

n= 4

2m-1=5

m=3

That is, 4th minima of 400 nm wave coincides with 3rd minima of 560 nm wave.

Location of this minima is-

y1 =[ (2×4-1).(1000).(400×10-6 ) ] / 2×0.1           (given,d=0.1 )

= 14 mm

Next 11th minima of 400 nm wave will coincides with 8thminima of 560 nm wave.

Location of this second minima is-

y2 = [(2×11-1).(1000).(400×10-6 )] /2×0.1

= 42 mm

So, the distance  between two sucessive regions of complete darkness is -

y2 - y1 = 42-14 = 28 mm

Answer 2

Answer:

For two successive regions of complete darkness, the minimum distance between them is equal to 28mm.

Explanation:

Given the distance between two slits $=0.1$

Consider that the $nth$ minima of $400nm$ coincides with $mth$ minima of $560nm$

$(2n-1)400=(2m-1)560\\ \frac{2n-1}{2m-1}=\frac{560}{400} =\frac{7}{5}$

$2n-1=7\\ 2n=8\\ n=4$     ,              $2m-1=5\\2m=6\\m=3$

Therefore, $4th$ minima of $400nm$ coincided with $3rd$ minima of $560nm$

$Y_{1}= \frac{7(1000)(400*10^{-6})}{2*0.1}=14mm$

Next, $11th$ minima of $400nm$ will coincide with $8th$ minima of 560nm

$Y_{2}= \frac{21(1000)(400*10^{-6})}{2*0.1}=42mm$

Minimum distance, $Y_{2}-Y_{1}=42-14=28mm$