Question

Two plane mirrors are facing each other. They are parallel, 3.00 cm apart, and 17.0 cm in length, as the drawing indicates. A laser beam is directed at the top mirror from the left edge of the bottom mirror. What is the smallest angle of incidence with respect to the top mirror, such that the laser beam (a) hits only one of the mirrors and (b) hits each mirror only once?

Answer 1

Thus, the smallest angle of incidence with respect to the top mirror, such that the laser beam hits only one of the mirrors = 70.6° to the normal

the smallest angle of incidence with respect to the top mirror, such that the laser beam hits each mirror only once = 62.2° to the normal

Step-by-step explanation:

If the laser beam is directed at the top mirror from the left edge of the bottom mirror; i.e the laser beam is shone from the left end of the plane; The we have:

`tan \theta = (3)/(17/2)\\\\` since the two plane mirror forma the 17.0 cm length and provided it hits only one mirror

`tan \theta = (3)/(8.5)`

`\theta = tan ^(-1) ((3)/(8.5))`

`\theta = 19.4^0`

Thus; i = (90° - 19.4°)

i = 70.6° to the normal

Thus, the smallest angle of incidence with respect to the top mirror, such that the laser beam hits only one of the mirrors = 70.6° to the normal

b)

`tan \theta = (3)/(17/3)\\\\` provided that the laser beam hits each mirror only once.

`\theta = tan ^(-1) ((3)/(5.7))`

`\theta = 27.8^0`

Thus; i = (90° - 27.8°)

i = 62.2° to the normal

Thus, the smallest angle of incidence with respect to the top mirror, such that the laser beam hits each mirror only once = 62.2° to the normal