Question

Consider a set if two slits each of width b = 5×10^−2 cm and separated by a distance d = 0.1 cm, illuminated by a monochromatic light of wavelength 6.328 × 10^−5 cm. If a convex lens of focal length 10 cm is placed beyond the double slit arrangement, calculate the positions of the minima inside the first diffraction minimum.

Answer 1

We know that the first minima

$d\sin \theta = (m+\frac{1}{2})\lambda$

We can write it as

$\sin \theta = \frac{\lambda}{d}(1+\frac{1}{2})$

Now, substituting the values

$\sin\theta = \frac{6.328 × 10^−5}{0.1\times 2}=3.164\times 10^{-4}$

here angle is very small hence we can write it as $\sin\theta\sim\theta\sim\tan\theta$

Hence, $\tan\theta = 3.164\times 10^{-4}$

position of the minima $x=f\tan\theta$

$x=10\times 3.164\times 10^{-4} =0.03164mm$

Now,

$\sin \theta = \frac{\lambda}{d}(1+\frac{1}{2})$

$\sin\theta = \frac{6.328 × 10^−5}{0.1}(1+\frac{1}{2})=\frac{6.328 × 10^−5}{0.1}(\frac{3}{2})$

$\sin \theta = 9.492\times 10^{-4}$

$\tan\theta\sim\sin\theta=9.492\times 10^{-4}$

Hence the position of minima

$x=f\tan\theta$

$x=10\times 9.492\times 10^{-4} =0.094mm$