Question

the distance between slits on a diffraction grating is 0.60 mm, and one of the angles of diffraction is 0.30°. the light forms a second - order bright band. how large is the path difference? 1047 nm 1571 nm 3142 nm 6284 nm

Explanation:

Step1: Recall the formula for diffraction grating

For a diffraction grating, the formula for bright bands is \(d\sin\theta = m\lambda\), where \(d\) is the distance between slits, \(\theta\) is the angle of diffraction, \(m\) is the order of the bright band, and \(\lambda\) is the wavelength (path difference).

Step2: Convert units

Given \(d = 0.60\space mm=0.60\times10^{- 3}\space m\), \(\theta = 0.30^{\circ}\), and \(m = 2\).

Step3: Calculate \(\sin\theta\)

\(\sin(0.30^{\circ})\approx0.00523\) (using a calculator).

Step4: Solve for \(\lambda\)

Substitute the values into the formula \(d\sin\theta = m\lambda\).
\(\lambda=\frac{d\sin\theta}{m}\)
\(\lambda=\frac{0.60\times 10^{-3}\times0.00523}{2}\space m\)
\(\lambda = 1.569\times10^{-6}\space m = 1569\space nm\) (This is incorrect due to wrong formula application. The path - difference for a diffraction grating bright band is \(m\lambda\). When \(m = 2\), we should use \(d\sin\theta=m\lambda\). But actually, the path - difference \(\Delta x = m\lambda\). Another way:
We know that for a diffraction grating, the condition for maxima is \(d\sin\theta=m\lambda\). The path - difference \(\Delta x=m\lambda\).
Given \(d = 0.60\space mm = 0.60\times10^{6}\space nm\), \(\theta = 0.30^{\circ}\), \(m = 2\)
\(\sin\theta\approx\frac{\theta}{57.3}\) (when \(\theta\) is in degrees, for small angles \(\sin\theta\approx\tan\theta\approx\theta\) (in radians), \(\theta = 0.30^{\circ}\), \(\theta\) (in radians)=\(\frac{0.30\pi}{180}\approx0.00523\) radians)
\(d\sin\theta=m\lambda\)
\(\lambda=\frac{d\sin\theta}{m}\)
\(d = 0.60\times10^{6}\space nm\), \(m = 2\), \(\sin\theta\approx0.00523\)
\(\lambda=\frac{0.60\times10^{6}\times0.00523}{2}=1569\space nm\) (wrong).
The correct formula for path - difference \(\Delta x = d\sin\theta\) (from the geometry of wave interference in diffraction grating).
\(d = 0.60\space mm=600000\space nm\), \(\theta = 0.30^{\circ}\), \(\sin\theta\approx0.00523\)
\(\Delta x=d\sin\theta=600000\times0.00523 = 3138\space nm\approx3142\space nm\)

Answer:

C. 3142 nm