Advertisements
Advertisements
Question
In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of nλ/a. Justify this by suitably dividing the slit to bring out the cancellation.
Solution
Consider that a single slit of width d is divided into n smaller slits.
∴ Width of each slit, `"d'" = "d"/"n"`
Angle of diffraction is given by the relation,
`theta = ("d"/"d'" lambda)/"d" = lambda/"d'"`
Now, each of these infinitesimally small slits sends zero intensity in directionθ. Hence, the combination of these slits will give zero intensity.
APPEARS IN
RELATED QUESTIONS
A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2.5 mm from the centre of the screen. Find the width of the slit.
Draw the intensity pattern for single slit diffraction.
Draw the intensity pattern for double slit interference.
State differences between interference and diffraction patterns.
(i) State the essential conditions for diffraction of light.
(ii) Explain diffraction of light due to a narrow single slit and the formation of pattern of fringes on the screen.
(iii) Find the relation for width of central maximum in terms of wavelength 'λ', width of slit 'a', and separation between slit and screen 'D'.
(iv) If the width of the slit is made double the original width, how does it affect the size and intensity of the central band?
Why cannot two independent monochromatic sources produce sustained interference pattern?
Deduce, with the help of Young's arrangement to produce interference pattern, an expression for the fringe width.
A parallel beam of light of 450 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1.5 m away. It is observed that the first minimum is at a distance of 3 mm from the centre of the screen. Calculate the width of the slit.
Derive the relation a sin θ = λ for the first minimum of the diffraction pattern produced due to a single slit of width ‘a’ using light of wavelength λ.
Using the monochromatic light of same wavelength in the experimental set-up of the diffraction pattern as well as in the interference pattern where the slit separation is 1 mm, 10 interference fringes are found to be within the central maximum of the diffraction pattern. Determine the width of the single slit, if the screen is kept at the same distance from the slit in the two cases.
Derive the relation a sin θ = λ for the first minimum of the diffraction pattern produced due to a single slit of width 'a' using light of wavelength λ.
