A fringe width of 6 mm was produced for two | vedclass

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

Question

(B) The formula for fringe width in Young's Double Slit Experiment is given by $\beta = \frac{\lambda D}{d}$.Here,$\beta = 6 \ mm = 6 	imes 10^{-3} \

Vedclass · Accepted Answer

$600$

Vedclass · Answer

(B) The formula for fringe width in Young's Double Slit Experiment is given by $\beta = \frac{\lambda D}{d}$.Here,$\beta = 6 \ mm = 6 	imes 10^{-3} \ m$,$d = 1 \ mm = 1 	imes 10^{-3} \ m$,and $D = 10 \ m$.Rearranging the formula to solve for wavelength $\lambda$:$\lambda = \frac{\beta d}{D} = \frac{(6 	imes 10^{-3} \ m) 	imes (1 	imes 10^{-3} \ m)}{10 \ m}$.$\lambda = \frac{6 	imes 10^{-6}}{10} \ m = 6 	imes 10^{-7} \ m$.To convert this to nanometers $(nm)$,we multiply by $10^9$:$\lambda = 6 	imes 10^{-7} 	imes 10^9 \ nm = 600 \ nm$.Thus,the value of $x$ is $600$.

$A$ fringe width of $6 \ mm$ was produced for two slits separated by $1 \ mm$ apart. The screen is placed $10 \ m$ away. The wavelength of light used is $'x' \ nm$. The value of $'x'$ to the nearest integer is

Similar Questions

In a Young's experiment,two coherent sources are placed $0.90 \, mm$ apart and the fringes are observed one metre away. If it produces the second dark fringe at a distance of $1 \, mm$ from the central fringe,the wavelength of monochromatic light used would be

$A$ graph is plotted between the fringe-width $(z)$ and the distance $(D)$ between the slit and the eye-piece,keeping other parameters constant. Which of the following graphs is correct?

In Young's double-slit experiment with a source of light of wavelength $6320 \ \mathring{A}$,the first maxima will occur when:

Who was the first to study the phenomenon of interference of light?

In $Y.D.S.E.$,how many maxima can be obtained on the screen if the wavelength of light used is $200 \ nm$ and $d = 700 \ nm$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module