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.

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.