Diffraction of light at a narrow opening
Practical Activity for 14-16
This is a simple and economic demonstration of the diffraction of light by a narrow opening, from which the wavelength of light can be determined.
Apparatus and Materials
- Diode laser (designed for educational use NOT a key chain laser) Class 2 either 635 mm or 670 mm
- Thin aluminium foil (0.1mm thickness available as a sealed wrapper of a milk powder tin)
- Stand with clamp
- Meter scale
- Sharp needle
- Travelling microscope
- Screen-white graph sheet pasted on cardboard
Health & Safety and Technical Notes
The power of a class 2 laser is less than 1 mW. This is not harmful even if it is seen directly: the blink response gives adequate protection. Warning: cheap laser pointers have not been tested and cannot be relied upon to be Class 2.
Keeping the foil over a smooth plane surface, make a small circular hole in it by pressing the needle tip gently into the foil. The rectangular slit can be prepared by using two straight aluminium strips over transparent sticky tape.
- Arrange the source (laser), aluminium foil with hole, and graph sheet pasted on cardboard as a screen, so that they lie in a straight line. Support the stands with the clamp as shown in the below diagram.
- Switch on the source and make fine adjustments so that the aperture in the foil is illuminated evenly.
- Move the screen about 1 m from the aluminium foil, so that the diffraction pattern is within the graph sheet pasted on the screen.
- The diffraction fringe pattern will be as shown above.
- For the calculation of the wavelength of the source used:
- Note down the diameter of the first minimum, Y , with the help of graph sheet markings.
- Note the distance between the foil and the screen, D.
- Using a travelling microscope, measure the diameter of the hole or slit width used, X.
- Calculate the wavelength of the light source, λ, using the formula λ = XY/ 2D.
- Let S1 S 2 be the diameter of the hole, acting like secondary coherent sources.
- Secondary wave fronts coming from these undergo superposition (interference), leading to bright and dark fringes on the screen.
- The ray diagram is shown above.
- MN is the path difference between S1 P’ & MP, where P is the centre of the first minimum and O is the centre of central maximum.
- Then MN = λ /2 ,…………………….(1)
- From triangle S1 NM, sinӨ = MN/S 1 M = MN / (X/2), or MN = (X/2) sinӨ……………………(2)
- From triangle OMP, sinӨ = PO /MP ≈Y1 / OM =Y1 / D………………………..(3),
- From (1),(2) &(3) Y 1 / D = λ / X, or λ = X (2Y 1 ) / 2D =XY/ 2D where the diameter of the first minimum, Y = 2Y 1 Hence λ =XY/ 2D
- Experimental results conducted at physics lab CMRIT BANGALORE:
- For circular hole, X = 0\. 018 cm, For rectangular slit, X = 0\. 021 cm
- D = 177 cm, D = 215 cm
- Measured Y =1\. 3 cm, Y =1\. 35 cm will give 659 nm< λ <661 nm
- This is similar to the λ calculated by a diffraction grating experiment, at 653 nm.
- Actual diffraction pattern recorded by 3.1 Mega pixel digital camera, in the dark room is shown above.
This experiment was submitted by Tukaram Shet, Senior Lecturer in Physics at CMRIT Bangalore