Light of wavelength 633nm passes through a single slit of width 1.50*10^-5 m. At what angle does the second interference minimum (m=2) occur? °

The second interference is minimum at angle 5 degree approximately

Explanation:

Given that the

Wavelength (λ) = 633nm

Width d = 1.5 × 10^-5m

n = 2

The wavelength (λ) of the light passing through a single slit is related to the angle Ø by

dsinØ = n(λ)

Substitutes all the parameters into the above formula

1.5×10^-5 × SinØ = 2 × 633 × 10^-9

Make SinØ the subject of formula

SinØ = 1.266×10^-6/1.5×10^-5

SinØ = 0.0844

Ø = Sin^-1( 0.0844 )

Ø = 4.84 degree.

The second interference is minimum at angle 5 degree approximately

d = 1.65 m

Explanation:

Given that,

The speed of a ball, v = 3 m/s

A ball rolls a level table that is 1.5 m above the floor.

We can find how long the ball is in free fall. We can use the second equation of kinematics as follows :

u is the initial speed in the vertical direction

So,

Now, using the formula of velocity.

So, the landing spot is at 1.65 m from the table.