The refractive index of a material changes by 0.014 as the colour of the light changes from red to violet. A rectangular slab of height 2.00 cm made of this material is placed on a newspaper. When viewed normally in yellow light, the letters appear 1.32 cm below the top surface of the slab. Calculate the dispersive power of the material.

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Solution

Given that : $μ - μ_{r}$ = 0.014

Again $μ_{y} = \frac{Real depth}{Apparent depth}$

$= \frac{2.00}{1.32} = 1.515$

So, dispersive power $= \frac{μ_{v} - μ_{r}}{μ_{y} - 1}$

$= \frac{0.014}{1.515 - 1}$

$= \frac{0.014}{0.515} = 0.027$

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Given that : μ−μr = 0.014 Again μy=Real depthApparent depth =2.001.32=1.515 So, dispersive power =μv−μrμy−1 =0.0141.515−1 =0.0140.515=0.027

Given that : $μ - μ_{r}$ = 0.014

Again $μ_{y} = \frac{Real depth}{Apparent depth}$

$= \frac{2.00}{1.32} = 1.515$

So, dispersive power $= \frac{μ_{v} - μ_{r}}{μ_{y} - 1}$

$= \frac{0.014}{1.515 - 1}$

$= \frac{0.014}{0.515} = 0.027$