In Young's double-slit experiment, the separation $d$ between the slits is $2 ext{ mm}$, the wavelength $\la

Question

In Young's double-slit experiment, the separation $d$ between the slits is $2 	ext{ mm}$, the wavelength $\lambda$ of the light used is $5896 	ext{ \AA}$ and distance $D$ between the screen and slits is $100 	ext{ cm}$. It is found that the angular width of the fringes is $0.20^{\circ}$. To increase the fringe angular width to $0.21^{\circ}$ (with same $\lambda$ and $D$) the separation between the slits needs to be changed to:

Accepted Answer

The angular width of fringes in Young's Double Slit Experiment (YDSE) is given by $	heta = \frac{\lambda}{d}$.
Given:
Initial angular width, $	heta_1 = 0.20^{\circ}$
Initial slit separation, $d_1 = 2 	ext{ mm}$
Final angular width, $	heta_2 = 0.21^{\circ}$
Let the final slit separation be $d_2$.
Since $\lambda$ is constant, we have $	heta \propto \frac{1}{d}$.
$\Rightarrow \frac{	heta_1}{	heta_2} = \frac{d_2}{d_1}$
$\Rightarrow d_2 = d_1 	imes \frac{	heta_1}{	heta_2}$
$\Rightarrow d_2 = 2 	imes \frac{0.20}{0.21} = \frac{40}{21} 	ext{ mm} \approx 1.9 	ext{ mm}$.
Thus, the new separation between the slits should be $1.9 	ext{ mm}$.

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