In finding out refractive index of glass slab, the following observations were made through travelling microscope 50 vernier scale division = 49 MSD; 20 divisions on main scale in each cm

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Solution

Q.
In finding out refractive index of glass slab, the following observations were made through travelling microscope 50 vernier scale division = 49 MSD; 20 divisions on main scale in each cm

For mark on paper
$M S R = 8.45 c m, V C = 26$

For mark on paper seen through slab
$M S R = 7.12 c m, V C = 41$

For powder particle on the top surface of the glass slab
$M S R = 4.05 c m, V C = 1$

(MSR = Main Scale Reading, VC = Vernier Coincidence)

Refractive index of the glass slab is : [2024]

1 1.24

2 1.35

3 1.42

4 1.52

Ans.
(3)

$MSD = \frac{1}{20} cm = 0.05 cm$

$50 VSD = 49 MSD \Rightarrow 1 VSD = \frac{49}{50} MSD$

$So, L . C = 1 MSD - 1 VSD = 1 MSD - \frac{49}{50} MSD$

$L . C = \frac{MSD}{50} = \frac{0.05 cm}{50} = 0.001 cm$

$t = 8.45 + 26 \times (0.001) = 8.476 cm$

$t' = 7.12 + 41 \times (0.001) = 7.161 cm$

$Width of powder particle = 4.05 + 1 \times (0.001) = 4.051 cm$

$t_{real} = 8.476 - 4.051 = 4.425 cm$

$t_{app} = 7.161 - 4.051 = 3.11 cm$

$μ_{glass} = \frac{t_{real}}{t_{app}} = \frac{4.425}{3.11} = 1.42$

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