Question

.3 In a Young’s double slit experiment 10 fringes are observed in a given segment of the screen

when light of wavelength 500.0 nm is used. When the experiment is performed in a medium

of refractive index 1.2, the number of fringes observed in the same segment remains 10. The

wavelength of light used in this case is​

Answer 1

number of fringe is inversely proportional to wavelength of monochromatic light in medium.

i.e., $N_1\lambda_1=N_2\lambda_2$

case 1: 10 fringes are observed when wavelength of monochromatic light 500nm is used in YDSE.

so, $N_1\lambda_1=10\times500$

case 2: 10 fringes are observed when YDSE performed in a medium of refractive index is 1.2.

Let wavelength of monochromatic light is $\lambda$

then, $N_2\lambda_2=N_2\frac{\lambda}{\mu}$ = 10 × $\lambda$/1.2

now, 10 × 500 = 10 × $\lambda$/1.2

or, 600 nm = $\lambda$

hence, answer is 600 nm

Answer 2

Answer:

600.0 nm

Explanation:

Wavelength of light used is inversely proportional to number of fringes observed.

∴λ₁ * n₁ = λ₂ * n₂     --> (1)

Here we have

       n₁ = 10;  λ₁ = 500 nm; μ = 1.2;  n₂ = 10

from eq. (1)

  10 * 500 =  10 *  λ /1.2       [ here, n =  λ/μ ]

  500 * 1.2 =  λ

   λ = 600.0 nm