Optical Phenomena

Problem : Find the position of the first minimum for a single slit of width 0.04 millimeters on a screen 2 meters distant, when light from a He-Ne laser λ = 632.8 nm is shone on the slit.

Problem : What is the irradiance at the position of the third maximum for a single slit of width 0.02 millimeters?

Solution for Problem 2 >>

The irradiance is given by I(θ) = I ₀ , where β = (Πd /λ)sinθ . It is given in the section lesson that the third maximum occurs when β = ±3.4707Π . Substituting this into the above we have: I = I ₀ = 0.082I ₀ . Thus the third maximum is only 8% as bright as each of the constituent waves.

Close

Problem : If we have a single slit 0.2 centimeters wide, a screen 1 meter distant, and the second maximum occurs at a position 1 centimeter along the screen, what must be the wavelength of light incident on the screen?

Problem : The Rayleigh Criterion for resolution states that two point sources are just resolved when the central maximum from one source falls on the first minimum of the diffraction pattern from the other source. If a car is approaching you at night with headlights 1 meter apart, how far away must you be in order to just resolve them? (treat the headlights as single slits of width 1 millimeter, and assume the lamps are monochromatic sodium sources of wavelength 589.29 nm).

Problem : A diffraction grating is a closely spaced array of apertures or obstacles forming a series of closely spaced slits. The simplest type, in which an incoming wavefront meets alternating opaque and transparent regions (with each opaque/transparent pair being of the same size as any other pair), is called a transmission grating. Determine the angular position of the maxima of such a grating in terms of λ and a , the distance between centers of adjacent slits. If light of 500 nm is incident of a slit containing 18920 slits and of width 5 centimeters, calculate the angular position of the second maximum.

Solution for Problem 5 >>

The analysis here is very much like that for Young's Double Slit. We assume that parallel beams of monochromatic light are incident on the slits, and that the slits are narrow enough such that diffraction causes light to spread over a very wide angle, such that interference can occur with all the other slits. Clearly, is the screen is very far off (compared to the width of the grating), all the beams travel approximately the same distance to the central point, so there is a maximum there. Constructive interference will also occur at angles θ where light from one slit must travel a distance mλ ( m integer) further than light from an adjacent slit. Thus if the distance between slits is a , this distance must be equal to a sinθ . Thus we can write out expression for the positions of the maxima as:

sinθ =

For the grating described, a will be equal to a = 0.05/18920 = 2.64×10^-6 . From the equation derived: θ ₂ = sin^-1 = 22.26 ^o .

Close