Question

The refractive index of a dense flint glass is 1.65 and for alcohol it is 1.36 with respect to air. Find the refractive index of alcohol with respect to glass.

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Answer 1

Given :

➔ Refractive index of glass = 1.65

➔ Refractive index of alcohol = 1.36

To Find :

➳ Refractive index of alcohol wrt glass.

Concept :

➠ Refractive index of medium is defined as the ratio of speed of light in vacuum to the speed of light in medium.

➠ It is an unitless as well as dimensionless quantity.

$\bigstar\:\underline{\boxed{\bf{n=\dfrac{c}{v}=\dfrac{speed\:of\:light\:in\:vacuum}{speed\:of\:light\:in\:medium}}}}$

➠ Refractive index of medium A wrt medium B is given by

$\bigstar\:\underline{\boxed{\bf{n_{AB}=\dfrac{n_A}{n_B}}}}$

Calculation :

⇒ $\tt\:n_{ag}=\dfrac{n_a}{n_g}$

⇒ $\tt\:n_{ag}=\dfrac{1.36}{1.65}$

⇒ $\underline{\boxed{\bf{n_{ag}=0.82}}}$

Answer 2

• Given -

• The refractive index of a dense Flint glass is 1.65 and for alcohol it is 1.36 with respect to air
• Refractive Index of dense Flint glass = 1.65
• Refractive Index of Alcohol = 1.36

• To Find -

• Refractive Index of Alcohol with respect to glass

• Solution -

• We know that, to find two substances refractive Index with respect to each other, we must divide them to get their refractive Index with respect to two substances which are given.
• Here,
• a = refractive Index of Alcohol
• g = refractive Index of dense Flint glass
• n = no. of substances
• Formula will become here -

★ $n(ag) = \frac{n(a)}{n(g)}$

Applying Values in given Conditions,

$⟹n(ag) = \frac{n(a)}{n(g)}$

$⟹n(ag) = \frac{1.36}{1.65}$

$⟹n(ag) = \frac{\cancel{1.36}}{\cancel{1.65}}$

$⟹n(ag) = 0.82$

⛬ Refractive Index of Alcohol with respect to dense Flint glass is 0.82