Question

In a Young's double slit experiment, 12 fringes are observed
to be formed in a certain segment of the screen when light of
wavelength 600 nm is used. If the wavelength of light is
changed to 400 nm, number of fringes observed in the same
segment of the screen is given by
​

Answer 1

we know, number of fringes = length of region/fringe width

also we know, fringe width, $\beta=\frac{\lambda D}{d}$

where D is distance between slit and screen , d is seperation between slits and $\lambda$ is wavelength of monochromatic light is used in YDSE.

so, number of fringes = $\frac{d\times\textbf{length of region}}{\lambda D}$

here it is clear that number of fringes is inversely proportional to wavelength of monochromatic light.

so, $\frac{N_1}{N_2}=\frac{\lambda_2}{\lambda_1}$

or, $\frac{12}{N_2}=\frac{400}{600}$

or, $N_2$ = 18

hence, number of fringes = 18

Answer 2

Answer:

The number of fringes observed in the same segment of the screen is 18.

Explanation:

Lets note down the given information from the question itself.

It is an Young's double slit experiment.

Number of fringes observed are 12.

Wavelength of light used previously is 600 nm.

The changed wavelength of light is 400 nm.

We know that,

number of fringes = length of the region ÷ fringe width.

We also know that,

Fringe width is ∝ to wavelength.

Therefore,

12 × 600 = x × 400

x = 18