Physics Questions and Answers

Refractive index
= \(\frac{Sin i}{Sin r}\) = \(\frac{\text{Vel. of light in air}}{\text{vel. of light in glass}}\)

= \(\frac{Sin i}{Sin 30^o}\)= \(\frac{3.0 \times 10^8}{2.0 \times 10^8}\)

= Sin i = \(\frac{(3.0 \times 10^8 \times Sin 30^o)}{2.0 \times 10^8}\)

= Sin i = \(\frac{(3.0 \times 10^8 \times 0.5)}{2.0 \times 10^8}\)

= \(\frac{1.5}{2}\)

= 0.75

No explanation has been provided for this answer.

The image of an object placed in front of a plane mirror is always as far behind the mirror as the object is in front of the mirror. Thus when the object is 4m in front of the mirror, the image will be 4m behind the mirror. But when the mirror is moved in towards the man, (that is the man is now 3m from the mirror), the image distance will again decrease by 1m from behind the mirror. Thus the new image position is 3m behind the mirror. Therefore, the distance between the man and his new image is 6m.

The lowest note (fundamental note) emitted by any stretched string has a fundamental frequecy fo = V/2l = 40Hz
The 1st overtone = 2fo = 2 x 40 = 80Hz
2nd overtone = 3fo = 3 x 40 = 120Hz
3rd overtone = 4fo = 4 x 40 = 160Hz
4th overtone = 5fo = 5 x 40 = 200Hz

Overtones between 40Hz and 180Hz are 80Hz, 120Hz, 160Hz = 3 Overtones

If a sound wave goes from a cold air region to a hot air region (which will be less dense), the velocity decreases, and since the frequency is always constant, the wave length decreases with the velocity