JEE MAIN - Physics (2023 - 29th January Morning Shift - No. 12)
In a Young's double slit experiment, two slits are illuminated with a light of wavelength $$800 \mathrm{~nm}$$. The line joining $$A_{1} P$$ is perpendicular to $$A_{1} A_{2}$$ as shown in the figure. If the first minimum is detected at $$P$$, the value of slits separation 'a' will be:
_29th_January_Morning_Shift_en_12_1.png)
The distance of screen from slits D = 5 cm
0.2 mm
0.5 mm
0.4 mm
0.1 mm
Explanation
$A_{2} P-A_{1} P=\frac{\lambda}{2} \quad$ (Condition of minima)
$$ \sqrt{\mathrm{D}^{2}+\mathrm{a}^{2}}-\mathrm{D}=\frac{\lambda}{2} $$
$$ \begin{aligned} & \mathrm{D}\left(1+\frac{\mathrm{a}^{2}}{\mathrm{D}^{2}}\right)^{1 / 2}-\mathrm{D}=\frac{\lambda}{2} \\\\ & \mathrm{D}\left(1+\frac{1}{2} \times \frac{\mathrm{a}^{2}}{\mathrm{D}^{2}}\right)-\mathrm{D}=\frac{\lambda}{2} \\\\ & \frac{\mathrm{a}^{2}}{2 \mathrm{D}}=\frac{\lambda}{2} \Rightarrow \mathrm{a}=\sqrt{\lambda \cdot \mathrm{D}} \\\\ & =\sqrt{800 \times 10^{-6} \times 50} \end{aligned} $$
$\mathrm{a}=0.2 \mathrm{~mm}$
$$ \sqrt{\mathrm{D}^{2}+\mathrm{a}^{2}}-\mathrm{D}=\frac{\lambda}{2} $$
$$ \begin{aligned} & \mathrm{D}\left(1+\frac{\mathrm{a}^{2}}{\mathrm{D}^{2}}\right)^{1 / 2}-\mathrm{D}=\frac{\lambda}{2} \\\\ & \mathrm{D}\left(1+\frac{1}{2} \times \frac{\mathrm{a}^{2}}{\mathrm{D}^{2}}\right)-\mathrm{D}=\frac{\lambda}{2} \\\\ & \frac{\mathrm{a}^{2}}{2 \mathrm{D}}=\frac{\lambda}{2} \Rightarrow \mathrm{a}=\sqrt{\lambda \cdot \mathrm{D}} \\\\ & =\sqrt{800 \times 10^{-6} \times 50} \end{aligned} $$
$\mathrm{a}=0.2 \mathrm{~mm}$
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