Question

The refractive index of a medium ' x ' with respect to ' y ' is 2/3 and the refractive index of the medium ' y ' with respect to medium ' z ' is 4/3 . Calculate the refractive index of the medium ' z ' with respect to medium ' x ' .. Find the speed of light in medium y if speed of light in medium x is 3*10^8 m/s

Answer 1

Answer:

The refractive index of medium ‘x’ with respect to medium ‘y’, nxy,or nx/ny = 2/3

The refractive index of medium ‘y’ with respect to medium ‘z’, nyz,or ny/nz = 4/3

To find: The refractive index of medium ‘z’ with respect to medium ‘x’, nzx.

Solution:

We know that the refractive index of medium 1 with respect to medium 2 is reciprocal to the refractive index of medium 2 with respect to medium 1.

Therefore,

The refractive index of medium ‘z’ with respect to medium ‘x’, nzx is given as-

nzx = nz/nx = nzy × nyz = nz/ny × ny/nx

Therefore,

nzx = 1/nyz × 1/nxy      (∵nzy=1/nyz)

nzx=1/ny/nz × 1/nx/ny

nzx = 1/4/3 × 1/2/3

nzx = 3/4 × 3/2

nzx = 9/8

Thus, the refractive index of medium ‘z’ with respect to medium ‘x’, nzx is 98.

Let the speed of light in medium ‘y’ be ‘V’, and the speed of light in medium ‘x’ be ‘C’ i.e. 3×108ms−1.

Therefore,

nyx, or ny/nx ⇒ 1/nxy ⇒1/nx/ny ⇒ 1/2/3 ⇒ 3/2

nyx, or ny/nx =3/2     (∵nyx=1/nxy)

y/x=3/2  ………………. (1)

c/V = 3×108/V    ………………. (2)

Equating (1) and (2) we get-

3/2 = 3× 108/V

3 × V = 2×3×108

3×V = 6×108

V = 6×108/3

V=2×108

Thus, the speed of light in the medium ‘y’ is 2×108.

Answer 2

Answer:

9/8

Explanation:

x/y=2/3

y/z=4/3

z/x=?

considering x,y,z as 3 different. Mediums, x is optically rarer than y and y is optically denser than z.

we know that if we multiply no. with reciprocals even in series we get 1

i.e 1/2*2/3*3/1(considering x/y, y/z and z/x)

=1
you can take other no.'s as well

now you see that x/y*y/z*z/x=1

*putting the given values*

2/3*4/3*z/x=1

8/9*z/x=1

z/x=1/8/9

therefore z/x =9/8

HENCE PROVED

SOLUTION BY IRAJ SHARMA