Question

If approximate refrctive index of sapphire is 1.8 then approximate decrease in velocity of light when light enters sapphire is

Answer 1

• Refractive index of sapphire=1.8
• Speed of light in sapphire=V=?

We know

$\boxed{\sf \mu=\dfrac{C}{V}}$

$\\ \sf\longmapsto 1.8=\dfrac{3\times 10^8}{V}$

$\\ \sf\longmapsto V=\dfrac{3\times 10^8}{1.8}$

$\\ \sf\longmapsto V=1.6\times 10^8m/s$

Answer 2

Answer:

45%

Explanation:

First, to make my life a lot simpler, 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.

$=(3*10^8)/1.8\\= (3*10^9)/ 18\\= (1*10^9) /6$

The percentage of the v is to c.

That is - v*100 /c

$=((10/6)/3) *100 (since-10^8-get-canceled-out)\\= (5/9) * 100\\= 0.55555 * 100$

= 55.555%

for the ease of mind, 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 for 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.

pls mark this brainliest if you think its justified. would mean a lot!

Hope it helped :)