As a reference, consider the line from the point perpendicular to the mirror.
That direction is called 'normal' to the mirror.
The ray on the right leaves the point traveling 5° to the right of the normal,
and leaves the mirror on a path that's 10° to the right of the normal.
The ray on the left leaves the point traveling 5° to the left of the normal,
and leaves the mirror on a path that's 10° to the left of the normal.
The angle between the two rays after they leave the mirror is 20° .
Frankly, Charlotte, if there were more than 5 points available for this answer,
I'd seriously consider giving you a drawing too.
The wavelength decreases to roughly half.
(The frequency roughly doubles.)
Answer:
E. 3h
Explanation:
We know that
u = 0 m/s.
velocity after t = 1s
v = u+gt = 0+9.81 x 1s= 9.81 m/s
distance covered in 1st sec
= =>> ut+0.5 x g x t²
=>>0 + 0.5x 9.81 x 1 = 4.90m
Let 4.90 be h
distance travelled in 2nd second will now be used
So velocity after t = 1s
=>>1 x t+ 0.5 x g x t²
=>9.81x 1 + 0.5 x 9.81 x 1 = 3 x 4.90
So since h= 4.90
Then the ans is 3x h = 3h