Question

The maximum number of possible interference for slit separation equal to 1.88 where is is the wavelength s of light used, in Young's Double slit experiments (a) Zero (6) 3 (6) infinite (d) 5 ) 3 () 5 5

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Answer 1

Answer:

D is correct answer

Explanation:

For possible interference maxima on the screen, the condition is

D sinθ=n λ ...(i)

Given : d= slit - width =2λ

∴2λsinθ=nλ

⇒2sinθ=n

The maximum value of sinθ is 1 , hence, n=2×1=2

Thus, Eq. (i) must be satisfied by 5 integer values ie, −2,−1,0,1,2. Hence, the maximum number of possible interference maxima is 5.

Answer 2

Explanation:

correct option is (infinite).

The condition for maxima in Young's double slit experiment is dsinθ=nλ where d is the separation between the slits and λ is the wavelength of light used.

The maxima that is farthest from the slits is infinitely up or infinitely down the screen and corresponds to θ=

2

π

.

So, with the given condition, we have, n=

λ

dsin(90°)

=

λ

d

=2

Thus, we have two maxima on the screen on either side of the central maxima. Thus, the maximum number of possible maxima observed are 2+1+2=5.