Question

In young’s experiment, the upper slit is covered by a thin glass plate of refractive index $\frac{4}{3}$ and of thickness $9\lambda$, where $\lambda$ is the wavelength of light used in the experiment. The lower slit is also covered by another glass plate of thickness $2\lambda$ and refractive index $\frac{3}{2}$ , as shown in figure. If $I_0$ is the intensity at point P due to slits $S_1$ and $S_2$ each, then :

• A. Intensity at point P is $4I_0$
• B. Two fringes have been shifted in upward direction after insertion of both the glass plates together.
• C. Optical path difference between the waves from $S_1$ and $S_2$ at point P is $2\lambda$.
• D. If the source S is shifted upwards by a small distance d2 then the fringe originally at P after inserting the plates, shifts downward by $D\left(\frac{d_2}{d_1}\right)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Intensity at point P is $4I_0$
B Two fringes have been shifted in upward direction after insertion of both the glass plates together.
C Optical path difference between the waves from $S_1$ and $S_2$ at point P is $2\lambda$.
D If the source S is shifted upwards by a small distance d2 then the fringe originally at P after inserting the plates, shifts downward by $D\left(\frac{d_2}{d_1}\right)$


Optical path difference due to
$S_{1}is=t({\mu }_1-1)=9\lambda (\frac{4}{3}-1)=3\lambda$
Optical path differ due to $S_2$ is
$=t({\mu }_2-1)=2\lambda (\frac{3}{2}-1)=\lambda$
Path diff. at point $P=3\lambda -\lambda =2\lambda$
Phase diff. $=\frac{2\pi }{\lambda }\times \lambda =4\lambda$
So net intensity at point P is
$I_{net}=I_0+I_0+2I_{0}cos4\pi I_{net}=4I_0$