Question

In YDSE using monochromatic visible light, the distance between the plane of slits and the screen is $1.7m$. At point $P$ on the screen which is directly in front of the upper slit, maximum path is observed. Now, the screen is moved $50cm$ closer to the plane of slits. Point $P$ now lies between third and fourth minima above the central maxima and the intensity at $P$ is one-fourth of the maximum intensity on the screen. Find the wavelength of light if the separation of slits is $2mm$.

• A. $2.9\times {10}^{-7}m$
• B. $3.9\times {10}^{-7}m$
• C. $5.9\times {10}^{-7}m$
• D. $6.9\times {10}^{-7}m$

Answer 1

The correct option is C $5.9\times {10}^{-7}m$

Since maxima is obtained in front of one of the slits, the difference at that point is a multiple of wavelength.

$\frac{\left(\frac{d}{2}\right)d}{D_1}=n\lambda$.....................Equation 1

here $D_1=1.7m$

When the distance between the two planes is reduced, the point lies between third and fourth minima above the central maxima.

Therefore, $\frac{5}{2}\lambda <\frac{\left(\frac{d}{2}\right)d}{D_2}<\frac{7}{2}\lambda$, here $D_2=1.7m-0.5m=1.2m$

$⟹\frac{5}{2}\lambda <n\lambda \frac{1.7}{1.2}<\frac{7}{2}\lambda$

Hence, $\frac{30}{17}<n<\frac{42}{17}$

$⟹n=2$

Using equation 1 we get,

$\lambda =5.9\times {10}^{-7}m$