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(D) Given: Fringe width $\beta = 0.03 \, cm = 3 	imes 10^{-4} \, m$,Distance $D = 1 \, m = 100 \, cm$,Focal length $f = 16 \, cm$,Image distance $v =

Vedclass · Accepted Answer

$6000 \, \mathring{A}$

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(D) Given: Fringe width $\beta = 0.03 \, cm = 3 	imes 10^{-4} \, m$,Distance $D = 1 \, m = 100 \, cm$,Focal length $f = 16 \, cm$,Image distance $v = 80 \, cm$,and distance between images $d' = 0.8 \, cm$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$,where $u$ is the distance of the slits from the lens:$\frac{1}{80} - \frac{1}{u} = \frac{1}{16} \Rightarrow \frac{1}{u} = \frac{1}{80} - \frac{1}{16} = \frac{1-5}{80} = -\frac{4}{80} = -\frac{1}{20}$.Thus,$u = -20 \, cm$ (the object is $20 \, cm$ from the lens).The magnification $m = \frac{v}{u} = \frac{80}{20} = 4$.Since $m = \frac{d'}{d}$,where $d$ is the distance between the slits:$4 = \frac{0.8}{d} \Rightarrow d = 0.2 \, cm = 2 	imes 10^{-3} \, m$.Using the fringe width formula $\beta = \frac{D \lambda}{d}$:$3 	imes 10^{-4} = \frac{1 	imes \lambda}{2 	imes 10^{-3}}$.$\lambda = 6 	imes 10^{-7} \, m = 6000 	imes 10^{-10} \, m = 6000 \, \mathring{A}$.

Similar Questions

$A$ double-slit experiment is performed with light of wavelength $500 \ nm$. If a thin plate of thickness $2 \ \mu m$ and refractive index $1.5$ is placed in front of one of the slits,the central fringe will shift by:

The fringe widths are found to be $\omega_1$ and $\omega_2$ respectively if a Young's double slit experiment is performed in media of refractive indices $n_1$ and $n_2$ respectively. The correct statement is:

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