"Two plane mirrors AB & CD each of length 2m are arranged parralel to each other 3^1/2m apart & a ray of light is incident at point 'A' of the mirror AB .How many reflections does the ray undergo? What is the distance travelled by the ray of light between the two mirrors ?what is the angle of deviation? " Please explain in detail.

Please refer to the figure below...

(a)

The number of reflections shown is three. First at point A, then at C and finally at B.

Now, we arrived at this number by calculating the horizontal distance travelled by the light ray between each successive reflection.

So, after it hits first at point A it reflects at angle 30 degrees (due to first law of reflection) and reaches point C'. As the two mirrors are parallel, the angle of incidence (and subsequently reflection) at this point C will also be 30 degrees (look at the figure above). 

Now, let us calculate the horizontal distance travelled by light ray (= AA' = CC') between the first and second reflections.

In triangle AC'A'

tan30 = AA' / A'C'

or

AA' = A'C'.tan30 = √3 x (1/√3)

thus,

AA' = 1m

so,

the light will travel the same horizontal distance (=1m) between the second reflection and third reflection (at C). Now as the lengths of the two mirrors are 2m each no further reflection will take pace after point C.

Thus, total number of reflections will be 3.

.

(b)

Now, the total distance travelled by light will be (look at figure above)

d = AC' + C'B

now, also due to symmetry

AC' = C'B 

now, in triangle AC'A'

sin30 = AA' / AC'

or

AC' = AA' / sin30 = 1/sin30

thus,

AC' = 2m

or

AC' = C'B = 2m 

so,

total distance travelled by light in between two mirrors will be

d = AC' + C'B = 2m + 2m

or

d = 4 m