Question

In Young's Double slit experiment, a thin glass mica strip of thickness $(t=15\lambda )$ is pasted in front of slit $S_1$. If the same strip is now shifted in front of other slit $S_2$, the number of fringes which cross the central point on the screen is
$(d>>D,\lambda <<dand{n}_{mica}=1.5)$

• A. 5
• B. 10
• C. 15
• D. 25

Answer 1

The correct option is C 15

When the mica sheet is placed anywhere in the path of the beam, the net path difference is given by,
$(\mu -1)t=n\lambda$
where $\mu$ is the refractive index of mica, $t$is the thickness, $n$ is the number of shifted fringes and $\lambda$ is the wavelength.
Given $\mu =1.5$, $t=15\lambda$
$⟹(1.5-1)15\lambda =n\lambda$
$⟹n=7.5$
If the number of fringes which will cross the central point on the screen is $N$, then
$N=2n=15$
Hence the total number of fringes shifted is $15$.