Sankalp NEET Full Test-4 Question-10 Solution

Question: 10: In a Young’s double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm , then the number of fringes he would observe in the same region of the screen is

(1) 9

(2) 12

(3) 6

(4) 8

Answer: Option (2)

Explanation:

In Young’s double slit experiment, the fringe width is given by

\beta = \frac{\lambda D}{d},

where \lambda is the wavelength of light, D is the distance between the slit and the screen, and d is the separation between the slits.

The number of fringes observed in a given fixed length on the screen is inversely proportional to the fringe width.

Hence, the number of fringes N is inversely proportional to the wavelength:

N \propto \frac{1}{\lambda}.

Let N_1 = 8 be the number of fringes corresponding to wavelength

\lambda_1 = 600 \text{ nm}, and

N_2 be the number of fringes for wavelength \lambda_2 = 400 \text{ nm}.

Using the proportionality,

\frac{N_2}{N_1} = \frac{\lambda_1}{\lambda_2}.

Substituting the values,

\frac{N_2}{8} = \frac{600}{400} = \frac{3}{2}.

N_2 = 8 \times \frac{3}{2} = 12.

Therefore, the number of fringes observed with wavelength 400 nm is 12.

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