Name: Diffraction Grating
Uploaded: 2019-07-19
Duration: 5 min 57 s
Description: A diffraction grating is an optical component with a periodic structure that splits and diffracts light into several beams travelling in different directions.

Question

In a double-slit experiment using light from a helium– neon laser, a student obtained the following results:

Width of $10$ Fringes $10 x = 1.5 cm$ Separation of slits $a = 1.0 mm$ Slit-to-screen distance $D = 2.40 m$ Determine the wavelength of the light. And

If The student moved the screen to a distance of $4.8 m$ From the slits. Determine the fringe separation $x$ Now.

Accepted Answer

Work out the fringe separation $x = \frac{1.5 \times 10^{- 2}}{10} = 1.5 \times 10^{- 3} m$

Substitute the values of $a, x$ and $D$ in the expression for wavelength $λ$ .

$λ = \frac{ax}{D}$

Therefore, $λ = \frac{1.0 \times 10^{- 3} \times 1.5 \times 10^{- 3}}{2.40}$ $= 6.3 \times 10^{- 7} m$

Wavelength of light used, $λ = 6.3 \times 10^{- 7} m$ And separation of slits, $a = 1.0 mm = 1.0 x 10^{- 3} m$ Remain the same. Separation of fringes. $x_{1} = 1.5 \times 10^{- 3} m$ , the screen moved to a distance $D_{2} = 4.8 m$ , $D_{1} = 2.4 m$

This means that $\frac{x_{1}}{D_{1}} = \frac{x_{2}}{D_{2}}$

Putting all the value we get $\frac{1.5 \times 10^{- 3}}{2.4} = \frac{x_{2}}{4.8}$

$∴ x_{2} = \frac{1.5 \times 10^{- 3} \times 4.8}{2.4} = 0.003 m = 3 mm$

Therefore, Doubling $D$ means $x$ Must also double, so separation of fringes $x_{2} = 3 mm$ .

David Sang and Graham Jones Solutions for Chapter: Superposition of Waves, Exercise 7: Question

David Sang Physics Solutions for Exercise - David Sang and Graham Jones Solutions for Chapter: Superposition of Waves, Exercise 7: Question

Questions from David Sang and Graham Jones Solutions for Chapter: Superposition of Waves, Exercise 7: Question with Hints & Solutions

Learn and Prepare for any exam you want !