Question

In a Young's double-slit experiment, the slit separation is $0.20$ cm and the slit to screen distance is $100cm$. If the wavelength of the source is $500nm$, the positions of the first three minima would be :

• A. $\pm 0.0125cm,\pm 0.0375cm,\pm 0.0625cm$
• B. $\pm 0.025cm,\pm 0.075cm,\pm 0.125cm$
• C. $\pm 12.5cm,\pm 37.5cm,\pm 62.5cm$
• D. $\pm 1.25cm,\pm 3.75cm,\pm 6.25cm$

Answer 1

Answer: (A)

$\pm 0.0125cm,\pm 0.0375cm,\pm 0.0625cm$

$\frac{dy}{D}=n\lambda$

$I^{st}minima=\frac{1\times 500\times {10}^{-9}}{2\times 0.2\times {10}^{-2}}$
$=1250\times {10}^{-7}$
$=\pm 0.125\times {10}^{-3}m$
$=0.0125cm$

$I{I}^{nd}minima=3\times I^{st}minima$
$=3\times 0.0125cm$
$=\pm 0.0375cm$

$II{I}^{rd}minima=5\times I^{st}minima$
$=5\times 0.0125cm$
$=\pm 0.0625cm.$