a narrow slit of width 0.3 mm is illuminated normally by a parallel beam of light of wavelength 6×10^-7 m.the Fraunhofer diffraction pattern is obtained on a screen placed in the focal plane of a convex lens of facal length 25. the lens is placed quite close close to the slit .find the width of central maximum on the screen .
(answer-1.0mm)
Answers
Answer:
The first and second maxima occur at α = 1.43π and 2.46π. Thus, Consequently the maxima will be separated on the screen by the distance given by (6.71– 3.9) × 10–3 × 70 = 0.2 cm. Read more on Sarthaks.com - https://www.sarthaks.com/423449/a-parallel-beam-of-light-is-incident-normally-on-narrow-slit-of-width-0-22-mm-the-fraunhofer?show=423451#a423451
Given info : a narrow slit of width 0.3 mm is illuminated normally by a parallel beam of light of wavelength 6×10^-7 m. the Fraunhofer diffraction pattern is obtained on a screen placed in the focal plane of a convex lens of facal length 25.
To find : the width of central maximum on the screen is ...
solution : using formula, width = 2λf/a
where a = width of slit = 0.3 mm = 0.3 × 10¯³ m
λ = wavelength of light = 6 × 10^-7 m
f = focal length of convex lens = 25 cm = 0.25 m
now, width = 2 × 6 × 10^-7 × 0.25/(0.3 × 10¯³)
= 10¯³ m
= 1 mm
Therefore the width of central maximum on the screen is 1mm.