Question 368: Wave Optics

Question

In Young's experiment, fringes are obtained on a screen placed at a distance $$75 \mathrm{~cm}$$ from the slits. When the separation between two narrow slits is doubled, then the fringe width is decreased. In order to obtain the initial fringe width, the screen should be moved through.

Accepted Answer

Fringe width $$\mathrm{X}=\frac{\lambda \mathrm{D}_1}{\mathrm{~d}_1}=\frac{\lambda \mathrm{D}_2}{\mathrm{~d}_2}$$
$$\begin{aligned}
& 	herefore \quad \frac{d_2}{d_1}=\frac{D_2}{D_1} \quad 	herefore 2=\frac{D_2}{D_1} \
& 	herefore \quad D_2=2 D_1=2 	imes 75=150 \mathrm{~cm}
\end{aligned}$$

Answer

$$150 \mathrm{~cm}$$ away from the slits.

Answer

$$75 \mathrm{~cm}$$ towards the slits.

Answer

$$75 \mathrm{~cm}$$ away from slits.

Answer

$$150 \mathrm{~cm}$$ towards the slits.

Wave Optics

In Young's experiment, fringes are obtained on a screen placed at a distance $75 \mathrm{~cm}$ from the slits. When the separation between two narrow slits is doubled, then the fringe width is decreased. In order to obtain the initial fringe width, the screen should be moved through.

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