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Question

(C) Let $\lambda$ be the wavelength of monochromatic light,$d$ be the distance between the coherent sources,and $D$ be the distance between the screen

Vedclass · Accepted Answer

Becomes $1/20$ times

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(C) Let $\lambda$ be the wavelength of monochromatic light,$d$ be the distance between the coherent sources,and $D$ be the distance between the screen and the source. The fringe width $\beta$ is given by:$\beta = \frac{D \lambda}{d}$Given the initial conditions: $d_1 = d$ and $D_1 = D$. The initial fringe width is $\beta_1 = \frac{D \lambda}{d}$.Given the new conditions: $d_2 = 10d$ and $D_2 = \frac{D}{2}$.The new fringe width $\beta_2$ is:$\beta_2 = \frac{D_2 \lambda}{d_2} = \frac{(\frac{D}{2}) \lambda}{10d} = \frac{D \lambda}{20d}$Comparing the new fringe width with the initial one:$\beta_2 = \frac{1}{20} \left( \frac{D \lambda}{d} \right) = \frac{\beta_1}{20}$Thus,the fringe width becomes $1/20$ times the original fringe width.

In a double slit experiment,the distance between slits is increased $10$ times whereas their distance from the screen is halved,then what is the fringe width?

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