Question

In a computer-based experiment to study diffraction, the width of the central diffraction peak is 15.20 mm. The wavelength of the laser is 638 ????m (1 ????m = 10-. m) and the distance from the screen containing the slits to the camera screen is 68.0 cm. Calculate the slit width ( make sure all your values are meters).Which equation should I use? slit width = (2)(638*10^-9)(0.68)/(15.20*10^-3) OR slit width = (1)(638*10^-9)(0.68)/(15.20*10^-3)? please explain why.

Answer 1

Answer:

$a = 5.7 \times 10^{-5} m$

Explanation:

As we know that position of first minimum on the either side of central maximum is given as

$a sin\theta = \lambda$

$\theta =sin^{-1} \frac{\lambda}{a}$

so the width of the central maximum is given as

$W = L (2\theta)$

so we have

$15.20 \times 10^{-3} = 0.68 \times 2(sin^{-1} \frac{\lambda}{a})$

so we have

$0.011 = sin^{-1} \frac{\lambda}{a}$

$0.011 = \frac{638 nm}{a}$

$a = 5.7 \times 10^{-5} m$