Optics 6 Question 40
42. A slit of width $d$ is placed in front of a lens of focal length $0.5 m$ and is illuminated normally with light of wavelength $5.89 \times 10^{-7} m$. The first diffraction minima on either side of the central diffraction maximum are separated by $2 \times 10^{-3} m$. The width $d$ of the slit is …… $m$.
(1997, 1M)
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Solution:
- Given
$$ \begin{aligned} 2 y & =2 \times 10^{-3} m \\ y & =1 \times 10^{-3} m \end{aligned} $$
First minima is obtained at
$$ \begin{aligned} d \sin \theta & =\lambda \text { but } \sin \theta \approx \tan \theta=\frac{y}{f} \\ \therefore \quad d \frac{y}{f} & =\lambda \\ d & =\frac{\lambda f}{y}=\frac{5.89 \times 10^{-7} \times 0.5}{1 \times 10^{-3}} \\ & =2.945 \times 10^{-4} m \end{aligned} $$
