a.) A single slit diffraction pattern is observed on a screen. The distance from
ID: 1446288 • Letter: A
Question
a.) A single slit diffraction pattern is observed on a screen. The distance from the slit to the screen is 0.8m. The distance from the central maximum to the first minimum is 0.9 cm. We know the wavelength of the laser light is 632nm. What is the width of the slit in mm?
b.) A single slit diffraction pattern is observed on a screen. The width of the slit is 0.02mm. The distance from the slit to the screen is 1m. If the wavelength of the laser light is 632.8nm. How far is the third minimum from the central maximum in cm?
c.) A single slit diffraction pattern is observed on a screen. The width of the slit is 0.02mm. The distance from the slit to the screen is 1m. If the wavelength of the laser light is 532nm. How far is the second minimum to the first minimum in cm? will thumbs up
Explanation / Answer
a) given
R = 0.8 m
y = 0.9 cm = 0.009 m
lamda = 632 nm
we know, for first dark fringe, y1 = lamda*R/d
d = lamda*R/y1
= 632*10^-9*0.8/0.009
= 5.61*10^-5 m
= 0.0561 mm
b) given
d = 0.02 mm
R = 1m
lamda = 632.8 nm
apply, y3 = 3*lamda*R/d
= 3*632.8*10^-9*1/(0.02*10^-3)
= 0.095 m
= 9.5 cm
c) given
d = 0.02 mm
R = 1m
lamda = 532 nm
apply, y2 = 2*lamda*R/d
= 2*532*10^-9*1/(0.02*10^-3)
= 0.0532 m
= 5.32 cm
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.