Question 357: Waves Optics

Question

In Young's double slit experiment, in an interference pattern, a minimum is observed exactly in front of one slit. The distance between the two coherent sources is '$$\mathrm{d}$$' and '$$\mathrm{D}$$' is the distance between the source and screen. The possible wavelengths used are inversely proportional to

Accepted Answer

If $$x$$ is the fringe width, then there will be a minimum in front of the slit if
$$\frac{\mathrm{d}}{2}=\frac{x}{2}, \frac{3 \mathrm{x}}{2}, \frac{5 \mathrm{x}}{2}, \ldots \ldots$$
or $$d=x, 3 x, 5 x, \ldots \ldots$$
$$	herefore x=d, \frac{d}{3}, \frac{d}{5}, \ldots \ldots$$
$$\lambda=\frac{\mathrm{xd}}{\mathrm{D}}$$
$$	herefore \lambda=\frac{\mathrm{d}^2}{\mathrm{D}}, \frac{\mathrm{d}^2}{3 \mathrm{D}}, \frac{\mathrm{d}^2}{5 \mathrm{D}}$$
$$	herefore \lambda$$ is inversely proportional to $$\mathrm{D}, 3 \mathrm{D}, 5 \mathrm{D}, \ldots$$.

Answer

D, 5D, 9D, ....

Answer

$$\mathrm{D}, 3 \mathrm{D}, 5 \mathrm{D}, \ldots$$

Answer

$$3 \mathrm{D}, 4 \mathrm{D}, 5 \mathrm{D}, \ldots$$

Answer

$$3 \mathrm{D}, 7 \mathrm{D}, 10 \mathrm{D}, \ldots$$

Waves Optics

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