Chapter 46: Problem 4

Question: Light of wavelength $\lambda~=~633~{\rm nm}$ is incident on a narrow slit. The angle between the first minimum on one side of the central maximum and the first minimum on the other side is $2\theta_1~=~1.97^o$. Find the width of the slit.

Solution: We know that the first minimum is located at an angle such that $a\sin\theta_1~=~\lambda$. Inserting the appropriate values gives

$$a~=~{633\times 10^{-9}\over \sin\left( {1.97\pi\over 2 \ 180}\right)}~=~3.68\times 10^{-5}~{\rm m} \ \ \ .$$