Question

A single slit has a width of 0.04 mm. A parallel beam of light of wavelength 560 nm
is incident normally on it. If the distance between the slit and the screen is 100 cm,
calculate the separation between the central maximum and the second minima in the
diffraction pattern

Answer 1

Answer on Question #73863, Physics / Optics

Question. A single slit has a width of $0.04 \, \text{mm}$. A parallel beam of light of wavelength $560 \, \text{nm}$ is incident normally on it. If the distance between the slit and the screen is $100 \, \text{cm}$, calculate the separation between the central maximum and the second minima in the diffraction pattern.

Given. $a = 0.04 \, \text{mm} = 0.04 \cdot 10^{-3} \, \text{m}; \lambda = 560 \, \text{nm} = 560 \cdot 10^{-9} \, \text{m}; l = 100 \, \text{cm} = 1 \, \text{m}; m = 2$.

Find. $b - ?$

Solution.

The condition of diffraction minima on a single slit

$$a \sin \varphi = \pm m \lambda \quad (m = 1, 2, \dots).$$

From the figure

$$b = l \operatorname{tg} \varphi.$$ $$b \ll l \rightarrow \operatorname{tg} \varphi \approx \sin \varphi \rightarrow b = l \sin \varphi \rightarrow \sin \varphi = \frac{b}{l}.$$

We get

$$a \cdot \frac{b}{l} = 2\lambda \rightarrow b = \frac{2 \cdot \lambda \cdot l}{a} = \frac{2 \cdot 560 \cdot 10^{-9} \cdot 1}{0.04 \cdot 10^{-3}} = 28 \cdot 10^{-3} \, \text{m} = 28 \, \text{mm}.$$

Answer. $b = 28 \, \text{mm}$

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