Question

In a standard YDSE setup, the fringe width on the screen is 1.5 mm. When a thin glass film is pasted in front of the upper slit, the fringe pattern shifts up. But it is seen that at a point P above central maxima where intensity was one fourth the intensity at central maxima, intensity remains the same. There were no maxima between central maxima and point P before film was introduced. What can be the thickness of the film? Take $\mu$ = 1.5, $\lambda$ = 450 nm :-

• A. 3 x ${10}^{-7}$m
• B. 5 x ${10}^{-7}$m
• C. 9 x ${10}^{-7}$m
• D. 1.4 x ${10}^{-6}$m

Answer 1

The correct option is D 1.4 x ${10}^{-6}$m

$\begin{array}{c}Pathdifferenceduetoslabshouldbeintegralmultipleof\lambda or\Delta x=n\lambda \\ \left(\mu -1\right)t=n\lambda n=\mathrm{1.2.3......}\\ ort=\frac{n\lambda }{\mu -1}\\ Forminimumvalueoft,wehaven=1\\ \therefore t=\frac{\lambda }{\mu -1}=\frac{\lambda }{\left(1.5-1\right)}=2\lambda \end{array}$

On putting the value of $\lambda$ we get

$t=1.4\times {10}^{-6}$

Hence, The option $D$ is the correct answer.