Answer:
The speed of proton when it emerges through the hole in the positive plate is .
Explanation:
Given that,
A parallel-plate capacitor is held at a potential difference of 250 V.
A A proton is fired toward a small hole in the negative plate with a speed of,
We need to find the speed when it emerges through the hole in the positive plate. It can be calculated using the conservation of energy as :
So, the speed of proton when it emerges through the hole in the positive plate is .
Answer:
the minimum wall thickness that will enhance the reflection of light is 146.9 nm
Explanation:
Given the data in the question;
At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.
At the second interface, no shift occurs,
condition for constructive interference;
t = ( m + 1/2) × λ/2n
where m = 0, 1, 2, 3 . . . . . .
now, the condition for the constructive interference;
t = mλ/2n
where t is the thickness of the soap bubble, λ is the wavelength of light and n is the refractive index of soap bubble.
so the minimum thickness of the film which will enhance reflection of light will be;
t = ( m + 1/2) × λ/2n
we substitute
t = ( 0 + 1/2) × 711 /2(1.21)
t = 0.5 × 711/2.42
t = 0.5 × 293.80165
t = 146.9 nm
Therefore, the minimum wall thickness that will enhance the reflection of light is 146.9 nm