Question

In Young?s experiment, the monochromatic light is used to illuminate two slits $A$ and $B$ as shown. Interference fringes are observed on a screen placed in front of the slits. Now if a thin glass plate is placed normally in the path of beam coming from the slit $A$, then

• A. fringe width will decrease
• B. fringes will disappear
• C. fringe width will increase
• D. there will be no change in fringe width

Answer 1

The Correct Option is D

Suppose $S_{1}$ and $S_{2}$ arc the slits at a distance $d$ from each other. Distance of screen be $D$. Let $P$ be a point where there is a bright fringe. A glass plate is placed in the path of the ray from $S_{1}$ to $P$. We know that the path difference between the rays in absence of glass plate is $\Delta x=S_{2} P-S_{1} P=\frac{d y}{D}$ In presence of the glass plate, the optical path length of the ray from $S_{1}$ to $P$ will be different. The total optical path length for this ray is given by $S_{1} P-t+\mu l$ $=S_{1} P+(\mu-1) t$ Where $\mu$ is the refractive index of the glass platc and $t$ is its thickness. Hence the new path difference is given by $\Delta x' =S_{2} P-\left[S_{1} P+(\mu-1) t\right]$ $=\Delta x-(\mu-1) t $ $=\frac{d y}{D}-(\mu-1) t$ For a bright fringe, $\Delta x'=n \lambda$ and $y=y n=$ distance of the bright fringe from the central fringe $\therefore \,\,\,\frac{d y_{u}}{D}-(\mu-1) t=n \lambda$ $\Rightarrow \,\,\,y_{n}=\frac{D}{d}[(\mu-1) t+n \lambda] $ $\therefore \,\,\,y_{n+1}-y_{n}=\omega=\frac{D \lambda}{d}$ Hence the fringe width remains constant.