Question

Chapter :- Light !!

Explanation Needed :)

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Answer 1

$\huge\sf\pink{Answer}$

☞ Refractive index of second medium is √2

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$\huge\sf\blue{Given}$

✭ sin C = 30°

✭ sin R = 90°

✭ Refractive index of medium 1 is 2√2

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$\huge\sf\gray{To \:Find}$

◈ Refractive Index of medium 2?

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$\huge\sf\purple{Steps}$

So here we may use Shell's Law, that is,

$\underline{\boxed{\sf sin \ C \times R_1 = sin \ R \times R_2}}$

• sin C = sin (Critical angle) = 30°
• sin R = sin (Reflected angle) = 90°
• R¹ = 2√2

Substituting the given values,

➝ $\sf sin \ C \times R_1 = sin \ R \times R_2$

➝ $\sf sin \ 30^{\circ} \times 2\sqrt{2} = sin \ 90^{\circ} \times R_2$

➝ $\sf \dfrac{1}{2} \times 2\sqrt{2} = 1 \times R_2$

➝ $\sf \orange{R_2 = \sqrt{2}}$

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☯ $\sf \underline{\sf Know \ More}$

»» Refraction is the bending of light from one medium to another

»» When a ray of light travels from a medium of less density to a medium of higher density it bends towards the normal

»» When a ray of light travels from a medium of higher density to a medium of lower density it bends away from the normal

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Answer 2

$\large{\underline{\sf Given:-}}$

• sin C = 30°
• sin R = 90°
• Refractive index of 1st medium = $2\sqrt{2}$

$\large{\underline{\sf Find:-}}$

• Refractive index of 2nd medium

$\large{\underline{\sf Solution:-}}$

we, know that

$\huge\underline{\boxed{ \sf n_{1} \times \sin C =n_{2} \times \sin R }}$

where,

• $\sf n_1 = 2\sqrt{2}$
• $\sf \sin C = {30}^{\circ}$
• $\sf \sin R = {90}^{\circ}$

So,

$\dashrightarrow\sf n_{1} \times \sin C =n_{2} \times \sin R$

$\dashrightarrow\sf 2 \sqrt{2} \times \sin {30}^{\circ}=n_{2} \times \sin {30}^{\circ}$

we, know

$\boxed{\begin{lgathered} \sin {30}^{ \circ} = \dfrac{1}{2} \\ \sin {90}^{ \circ} = 1\end{lgathered}}$

So,

$\dashrightarrow\sf 2 \sqrt{2} \times \dfrac{1}{2} =n_{2} \times 1$

$\dashrightarrow\sf \sqrt{2} \times 1 =n_{2}$

$\dashrightarrow\sf \sqrt{2}=n_{2}$

$\huge{\therefore \pink{\sf n_{2} =\sqrt{2}}}$

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Hence, Refractive Index of second 2nd medium = $2\sqrt{2}$