Question

if approximate refractive index of sapphire is 1.8 then approximate decrease in velocity of light,when light enters sapphire is________
1)45%
2)45%
3)50%
4)55%​

Answer 1

Explanation:

45% is enters sapphire is

Answer 2

Answer:

45%

Explanation:

First, we should calculate the speed of light in Sapphire.

So, given-

         speed of light in air = 3 * 10^8 m/s = c

         refractive index of sapphire = 1.8 = Ri(s)

         speed of light in sapphire = v

we know- refractive index = speed of light in air / speed of light in medium

so- Ri(s) = speed of light in air / refractive index.

= (3*10^8)/1/8

= (30*10^8)/18

= (10*10^8)/6 = (10/6)*10^8

The percentage of the v is to c, That is - v*100 /c -

= ((10/6) / 3)*100 (here, 10^8 gets canceled out)

= (10/18) * 100

= (5/9) *100

= 55.555%

for simpler calculations, let us take this as 55%

since this is how much more v is to c, the decrease can be written as -

= 100% - 55%

= 45%

Therefore, there has been a decrease in the speed of light in sapphire from air by about 45%.

I know this can be simplified a lot, but I think this was a bit confusing and misleading, so I did it the long way.