In finding out the refractive index of a glas | vedclass

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(A) $1 	ext{ MSD} = \frac{1 	ext{ cm}}{20} = 0.05 	ext{ cm}$.$1 	ext{ VSD} = \frac{49}{50} 	ext{ MSD} = \frac{49}{50} 	imes 0.05 	ext{ cm} = 0.

Vedclass · Accepted Answer

$1.42$

Vedclass · Answer

(A) $1 	ext{ MSD} = \frac{1 	ext{ cm}}{20} = 0.05 	ext{ cm}$.$1 	ext{ VSD} = \frac{49}{50} 	ext{ MSD} = \frac{49}{50} 	imes 0.05 	ext{ cm} = 0.049 	ext{ cm}$.$LC = 1 	ext{ MSD} - 1 	ext{ VSD} = 0.05 - 0.049 = 0.001 	ext{ cm}$.For the mark on paper,$L_1 = 8.45 	ext{ cm} + 26 	imes 0.001 	ext{ cm} = 8.476 	ext{ cm} = 84.76 	ext{ mm}$.For the mark on paper through the slab,$L_2 = 7.12 	ext{ cm} + 41 	imes 0.001 	ext{ cm} = 7.161 	ext{ cm} = 71.61 	ext{ mm}$.For the powder particle on the top surface,$ZE = 4.05 	ext{ cm} + 1 	imes 0.001 	ext{ cm} = 4.051 	ext{ cm} = 40.51 	ext{ mm}$.Actual $L_1 = 84.76 - 40.51 = 44.25 	ext{ mm}$.Actual $L_2 = 71.61 - 40.51 = 31.10 	ext{ mm}$.Since $\mu = \frac{	ext{Real Depth}}{	ext{Apparent Depth}} = \frac{L_1}{L_2}$,$\mu = \frac{44.25}{31.10} \approx 1.42$.

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