Using monochromatic light of wavelength lambd | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the thickness of the mica sheet $t = 1.8 	imes 10^{-6} \ m$,refractive index $\mu = 1.6$,and the shift $y = (\mu - 1)t \frac{D}{d}$.The fri

Vedclass · Accepted Answer

$540$

Vedclass · Answer

(B) Given the thickness of the mica sheet $t = 1.8 	imes 10^{-6} \ m$,refractive index $\mu = 1.6$,and the shift $y = (\mu - 1)t \frac{D}{d}$.The fringe width in the first case is $\beta = \frac{\lambda D}{d}$.The fringe width in the third case is $\beta' = \frac{\lambda (2D)}{d} = 2\beta$.Given the condition $y = \beta'$,we have $(\mu - 1)t \frac{D}{d} = 2 \frac{\lambda D}{d}$.Simplifying,we get $(\mu - 1)t = 2\lambda$.Substituting the values: $(1.6 - 1) 	imes 1.8 	imes 10^{-6} = 2\lambda$.$0.6 	imes 1.8 	imes 10^{-6} = 2\lambda$.$1.08 	imes 10^{-6} = 2\lambda$.$\lambda = 0.54 	imes 10^{-6} \ m = 540 \ nm$.

Using monochromatic light of wavelength $\lambda$,an experimentalist sets up the Young's double slit experiment in three ways as shown. If she observes that $y = \beta'$,the wavelength of light used is.....$nm$

Similar Questions

The intensity at the maximum in a Young's double slit experiment is $I_0$. The distance between two slits is $d = 5\lambda$,where $\lambda$ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance $D = 10d$?

In an interference pattern,if the ratio of slit widths is $1:9$,find the ratio of maximum to minimum intensity.

In Young's double-slit experiment,the intensity at a point on the screen is $I_0/4$. Find the angular position of this point.

The correct curve between fringe width $\beta$ and the distance between the slits $(d)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module