Chapter 33: Q. 40 (page 956)

Helium atoms emit light at several wavelengths. Light from a helium lamp illuminates a diffraction grating and is observed on a screen $50.00 c m$ behind the grating. The emission at wavelength $501.5 n m$ creates a first-order bright fringe $21.90 c m$ from the central maximum. What is the wavelength of the bright fringe that is $31.60 c m$ from the central maximum?

Short Answer

Expert verified

The luminous fringe from the central maxima has a wavelength, $λ = 667.8 nm .$

Step by step solution

Step: 1 Finding the space grating:

Equation of grating by

$d \sin (θ_{m}) = m λ$

Grating space $d$ is constant as

$\frac{m_{a} λ_{a}}{\sin (θ_{m, a})} = d = \frac{m_{b} λ_{b}}{\sin (θ_{m, b})} .$

Step: 2 Equating equation:

We have to find $\sin (θ_{m, a}), \sin (θ_{m, b})$ as easy as follows by

role="math" localid="1649090960925" $λ_{b} = \frac{m_{a} λ_{a} \sin (θ_{m, b})}{m_{b} \sin (θ_{m, a})} Y_{x} = L \tan (θ_{m, x}) \Rightarrow θ_{m, x} = \tan^{- 1} (\frac{Y_{x}}{L}) .$

Step: 3 Calculating wavellength:

We may proceed to identify the unknown wavelength for our numerical example after being told explicitly that $m_{a} = 1$ and assuming that $m_{b} = 1$ as well (since this issue would have an infinite number of solutions if m b were not determined):

localid="1649156207047" $\begin{array}{l} λ_{b} = \frac{1 \times 5.015 \times 10^{- 7} \times \sin (\tan^{- 1} (\frac{0.316}{0.5}))}{1 \times \sin (\tan^{- 1} (\frac{0.219}{0.5}))} \\ \begin{array}{l} λ_{b} = = \frac{5.015 \times 10^{- 7} \times \sin ({32.29}^{\circ})}{\sin ({23.65}^{\circ})} \\ λ = 667.8 nm . \end{array} \end{array}$