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Two slits are illuminated with green light (λ = 540 nm). The slits are 0.05 mm apart and the distance to the screen is 1.5 m. At what distan
Question
Two slits are illuminated with green light (λ = 540 nm). The slits are 0.05 mm apart and the distance to the screen is 1.5 m. At what distance (in mm) from the central maximum on the screen is the average intensity 50% of the intensity of the central maximum?
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Physics
3 years
2021-09-03T21:28:43+00:00
2021-09-03T21:28:43+00:00 1 Answers
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Answers ( )
Answer:
0.21486 mm
Explanation:
The formula for the maximum intensity is given by;
I = I_o•cos²(Φ/2)
Now,we are not given Φ but it can be expressed in terms of what we are given as; Φ = πdy/(λL)
Where;
y is the distance from the central maximum
d is the distance between the slits
λ is the wavelength
L is the distance to the screen
Thus;
I = I_o•πdy/(λL)
We are given;
d = 0.05 mm = 0.5 × 10^(-3) m
λ = 540 nm = 540 × 10^(-9) m
L = 1.25 m
I/I_o = 50% = 0.5
From earlier, we saw that;
I = I_o•πdy/(λL)
We have I/I_o = 0.5
Thus;
I/I_o = πdy/(λL)
Plugging in the relevant values;
0.5 = (π × 0.5 × 10^(-3) × y)/(540 × 10^(-9) × 1.25)
Making y the subject, we have;
y = (0.5 × 540 × 10^(-9) × 1.25)/(π × 0.5 × 10^(-3))
y = 0.00021486 m
Converting to mm, we have;
y = 0.21486 mm