Question

Two slits spaced 0.260 mm apart are placed 0.700 m from a screen and illuminated by coherent light with a wavelength of 660 nm. The intensity at the center of the central maximum is ( theta=0) (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to Io/2?​

Answer 1

Answer:

Let x be the minimum distance from the central maximum then,

2

I

o

=I

o

cos

2

(

β

πx

)

β

πx

=

4

π

⇒x=

4d

λD

=

4×0.25×10

−3

6.5×10

−7

×0.75

=4.875×10

−4

m≈0.5mm

Answer 2

Explanation:

Given Two slits spaced 0.260 mm apart are placed 0.700 m from a screen and illuminated by coherent light with a wavelength of 660 nm. The intensity at the center of the central maximum is ( theta=0) (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to Io/2?

• Now we have
•       d = 0.26 mm = 0.26 x 10^-3 m
•       R = 0.70 m
•      λ = 660 mm = 660 x 10^-9 m
•     I = Io / 2
• So we have after the central maxima the angle of dark fringe m = 0 will be
•                   d sin theta = (m + ½ )λ
•                                      = (0 + ½ )λ
• Or sin theta = ½ λ / d
•            theta = sin^-1 (1/2 λ / d)
•                      = sin^-1 (1/2 x 660 x 10^-9 / 0.26 x 10^-3)
•                      = sin^-1 (1269.23 x 10^-6)
•                     =  sin^-1 0.001269
•                     = 0.0727 degree
• Now distance on the screen between the first dark fringe and central maxima will be
•                           z = R tan theta
•                           z = 0.70 x tan 0.0727
•                           z = 8.881 x 10^-4 m
• Now intensity as a function will be
•                      I = Io cos^2 (π dy / λR)
•                    Io / 2 = Io cos^2 (π dy / λR)
•            So we get
•                    ½ = cos^2 (π dy / λR)
•              Let  (π dy / λR) = x
•            cos^2 x = ½  
•                 x = cos^-1 √1/2  
• By converting into radians we get
•                    x = ¼ π  
•       (π dy / λR) = ¼ π
• Or  (dy / λR) = 1/4
• Or y = λR / 4d
• Substituting the values we get
•     y = 660 x 10^-9 x 0.70 / 4 x 0.26 x 10^-3
•     y = 4.442 x 10^-4 m

Reference link will be

https://brainly.in/question/2809614