Question 26

Consider a Young's double slit experiment as shown in figure. What should be the slit separation $$d$$ in terms of wavelength $$\lambda$$ such that the first minima occurs directly in front of the slit ($$S_1$$)?

$$S_1S_2 = d, \quad D = 2d$$

$$S_1P = D = 2d$$

$$S_2P = \sqrt{D^2 + d^2} = \sqrt{(2d)^2 + d^2} = \sqrt{5d^2} = \sqrt{5}d$$

$$\Delta x = S_2P - S_1P = \sqrt{5}d - 2d = (\sqrt{5} - 2)d$$

Condition for the first minima: $$\Delta x = \frac{\lambda}{2}$$

$$(\sqrt{5} - 2)d = \frac{\lambda}{2} \implies d = \frac{\lambda}{2(\sqrt{5} - 2)}$$

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