A monochromatic beam of light is travelling from medium A of refractive index n1 to a medium B of refractive index n2. In the medium A, there are x number of waves in certain distance. In the medium B, there are y number of waves in the same distance. Then, refractive index of medium A with respect to medium B is

Question

A monochromatic beam of light is travelling from medium A of refractive index n1 to a medium B of refractive index n2. In the medium A, there are x number of waves in certain distance. In the medium B, there are y number of waves in the same distance. Then, refractive index of medium A with respect to medium B is (A)  (y/x) (B) √(x/y) (C)  (x/y-x) (D)  (x/y)

Tardigrade Admin · Accepted Answer

If x number of waves of wavelength λ1 and y number of waves of wavelength λ2 are present in same distance s, then s=x λ1=y λ2 or (x/y)=(λ2/λ1)=(n1/n2) (n1/n2)=(x/y)

$\frac{y}{x}$

$\frac{x}{y}$

$\frac{x}{y - x}$

$\frac{x}{y}$

If x number of waves of wavelength $λ_{1}$ and y number of waves of wavelength $λ_{2}$ are present in same distance s, then
$s = x λ_{1} = y λ_{2}$
or $\frac{x}{y} = \frac{λ _{2}}{λ _{1}} = \frac{n _{1}}{n _{2}}$
$\frac{n _{1}}{n _{2}} = \frac{x}{y}$

xy​

yx​​

y−xx​

yx​

If x number of waves of wavelength λ1​ and y number of waves of wavelength λ2​ are present in same distance s, then s=xλ1​=yλ2​ or yx​=λ1​λ2​​=n2​n1​​ n2​n1​​=yx​

$\frac{y}{x}$

$\frac{x}{y}$

$\frac{x}{y - x}$

$\frac{x}{y}$

If x number of waves of wavelength $λ_{1}$ and y number of waves of wavelength $λ_{2}$ are present in same distance s, then
$s = x λ_{1} = y λ_{2}$
or $\frac{x}{y} = \frac{λ _{2}}{λ _{1}} = \frac{n _{1}}{n _{2}}$
$\frac{n _{1}}{n _{2}} = \frac{x}{y}$