Question

A point source is fixed 1.0 m away from a large screen. Call the line normal to the screen surface and passing through the center of the point source the z axis. When a sheet of cardboard in which a square hole 0.020 m on a side has been cut is placed between the point source and the screen, 0.50 m from the point source with the hole centered on the z axis, a bright square shows up on the screen. If, instead, a second sheet of cardboard with a similar square hole is placed between the point source and screen, 0.25 m from the point source with the hole centered on the z axis, the bright square it casts on the screen is identical to the bright square from the first sheet. What is the length of the side of the hole in this sheet?

Answer 1

The length of the side of the hole in the second cardboard sheet is
`L_2 = 0.01m`

Step-by-step explanation:

From the question we are told that

The distance of the point source from the screen is
`d = 1.0 m`

The length of a side of the first square hole is
`L_1 = 0.020 \ m`

The distance of the cardboard from the point source is
`D_1 = 0.50\ m`

The distance of the second cardboard from the point source is
`D_2 = 0.25 \ m`

Let take the
`\alpha_(max )` as the angle at which the light is passing through the edges of the cardboards square hole

Since the bright square casted on the screen by both square holes on the individual cardboards are then it means that

`\alpha_(max) __(1)} = \alpha_(max) __(2)}`

This implies that

`tan (\alpha_(max) __(1)}) = tan (\alpha_(max) __(2)})`

Looking at this from the SOHCAHTOA concept

`tan (\alpha_(max) __(1)}) = (opposite)/(Adjacent)`

Here opposite is the length of the side of the first cardboard square hole

and

Adjacent is the distance of the from the first cardboard square hole to the point source

And for

`tan (\alpha_(max) __(2)}) = (opposite)/(Adjacent)`

Here opposite is the length of the side of the second cardboards square hole (let denote it with
`L_2`)

and

Adjacent is the distance of the from the second cardboards square hole to the point source

So

`tan (\alpha_(max) __(1)}) = (0.020)/(0.50)`

And

`tan (\alpha_(max) __(2)}) = (L_2)/(0.25)`

Substituting this into the above equation

`(0.020)/(0.50) = (L_2)/(0.25)`

Making
`L_2` the subject

`L_2 = (0.25 *0.020)/(0.50)`

`L_2 = 0.01m`

Since it is a square hole the sides are the same hence

The length of the side of the hole in the second cardboard sheet is
`L_2 = 0.01m`