Question

a diffraction grading of width w produces a deviation θ in second order for light of wavelength λ. the total number n of slits in the grating is given by
○ \\(\frac{w}{λ}\sin θ\\)
○ \\(\frac{w}{2λ}\sin θ\\)
○ \\(\frac{λw}{2}\sin θ\\)
○ \\(\frac{2wλ}{\sinθ}\\)
○ \\(\frac{2λ}{\sinθ}\\)

Explanation:

Step1: Recall Diffraction Grating Formula

The formula for a diffraction grating is \( d\sin\theta = n\lambda \), where \( d \) is the slit separation, \( n \) is the order, \( \lambda \) is the wavelength, and \( \theta \) is the deviation angle. Here, \( n = 2 \) (second order). Also, the total width \( W = N \times d \), so \( d=\frac{W}{N} \).

Step2: Substitute \( d \) into Grating Formula

Substitute \( d=\frac{W}{N} \) into \( d\sin\theta = n\lambda \):
\( \frac{W}{N}\sin\theta = 2\lambda \) (since \( n = 2 \)).

Step3: Solve for \( N \)

Rearrange the equation to solve for \( N \):
\( N\sin\theta = \frac{W \times 2\lambda}{\sin\theta} \)? Wait, no—wait, from \( \frac{W}{N}\sin\theta = 2\lambda \), multiply both sides by \( N \): \( W\sin\theta = 2\lambda N \), then divide both sides by \( 2\lambda \):
Wait, no, let's redo. From \( \frac{W}{N}\sin\theta = 2\lambda \), solve for \( N \):
Multiply both sides by \( N \): \( W\sin\theta = 2\lambda N \)
Then divide both sides by \( 2\lambda \): \( N=\frac{W\sin\theta}{2\lambda} \)? Wait, no—wait, the options: let's check the options again. Wait, maybe I mixed up. Wait, \( d = \frac{W}{N} \), so \( \frac{W}{N}\sin\theta = n\lambda \). For \( n = 2 \), \( \frac{W}{N}\sin\theta = 2\lambda \). Then \( N = \frac{W\sin\theta}{2\lambda} \)? Wait, no, the options have \( \frac{W}{2\lambda}\sin\theta \), which is the same as \( \frac{W\sin\theta}{2\lambda} \). Wait, the second option is \( \frac{W}{2\lambda}\sin\theta \), which matches \( N = \frac{W\sin\theta}{2\lambda} \). Wait, let's check the options:

Option 2: \( \frac{W}{2\lambda}\sin\theta \), which is \( N = \frac{W\sin\theta}{2\lambda} \), which comes from \( d\sin\theta = n\lambda \) and \( d = W/N \), so \( (W/N)\sin\theta = 2\lambda \implies N = \frac{W\sin\theta}{2\lambda} \), which is \( \frac{W}{2\lambda}\sin\theta \).

Answer:

\(\boldsymbol{\frac{W}{2\lambda}\sin\theta}\) (the second option: \(\frac{W}{2\lambda}\sin\theta\))