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प्रश्न
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.
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उत्तर
The distance of the nth minimum from the center of the screen is, `x_n =(nDlambda)/a`
Where, D = distance of slit from screen, λ = wavelength of the light, a = width of the slit
For first minimum, n = 1:
`3 xx 10^-3 = (1xx (1.5) xx (450 xx 10^-9))/a => a = (1 xx (1.5) xx (450 xx 10^-9))/(3 xx 10^-3)`
∴ a = 0.225 mm
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संबंधित प्रश्न
Draw the intensity pattern for double slit interference.
(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?
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.
Which items are used to form the narrow, uniform slit?
Which electric bulb is required for the observation?
How should the slit be oriented relative to the filament of the lit glass bulb?
In what order are the colour filters used when repeating the observation?
What appears when the slit is aligned parallel to the filament?
Why is the fringe pattern wider with a red colour filter?
