A student is trying to make an
accurate measurement of the wavelength of green light from a mercury lamp alpha
= 546 nm). Using a double slit of separation of 0.50 mm, he finds he can see
ten clear fringes on a screen at a distance of 0.80 m from the slits. He then
tries an alternative experiment using a diffraction grating that has 3000
lines/cm.
(i) What will be the width of the
ten fringes that he can measure in the first experiment?
(ii) What will be the angle of
the second-order maximum in the second experiment?
(iii) Suggest which experiment
you think will give the more accurate measurement of wavelength.
Question Collected from Telegram Group.
Solution:
(i)Given:
Wavelength (λ) = 546nm = 546 ×
10-9m
Distance between slits and screen
(D) =0.8m
Separation between Slits (d) =
0.50mm =0.5×
10-3m
We Know:
Width(β) = $\frac{{n\lambda D}}{d}$
β $ = \frac{{10 \times 546
\times {{10}^{ - 9}} \times 0.8}}{{0.5 \times {{10}^{ - 3}}}}$
β=8.736×10-3
(ii) Given,
Number of lines per cm (N) = 3000lines/cm = 300000lines/m
Grating Element (a+b) = $\frac{1}{N}$=$\frac{1}{300000}$=3.33×10-6
Wavelength (λ) = 546nm = 546 ×
10-9m
Order (n) = 2
We Know:
(a+b)Sin θ = n λ
(3.33×10-6) Sin θ = 2×546
×
10-9
On Solving,
Sin θ = 0.327
θ=19.14˚
Thus, Angle of second order maxima is θ=19.14˚.
(iii) The second experiment will give the more accurate measurement of wavelength because the diffraction grating has a much higher slit separation than the double slit. This means that the fringes in the diffraction pattern will be much narrower, which will make them easier to measure accurately.