Question

A glass plate of thickness $12\times {10}^{-3}mm$ is placed in the path of one of the interfering beams in Young's double slit arrangement. Light of wavelength $6000A^o$ is used in the arrangement. If the central band is shifted by a distance equal to 40times the width of bands then the refractive index of glass will be

• A. $\frac{5}{4}$
• B. $\frac{4}{3}$
• C. $\frac{3}{2}$
• D. $\frac{2}{1}$

Answer 1

The correct option is D $\frac{2}{1}$

Width of band is consecutive distance between maxima and minima, which is half of fringe width
As the initial path difference was zero at center, the after placing the glass plate the path difference will be
$\Delta x=(\mu -1)t-\frac{yd}{D}$
for first maxima, path difference must be zero
$(\mu -1)t-\frac{yd}{D}=0$
given $y=40\frac{\beta }{2}=\frac{40\lambda D}{2d}$
$(\mu -1)t-\frac{20\frac{\lambda D}{d}\times d}{D}=0$
$(\mu -1)t=20\lambda$
$(\mu -1)=\frac{20\times 6000\times {10}^{-10}}{12\times {10}^{-6}}$
$(\mu -1)=1\Rightarrow \mu =2$