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Byju's Answer

Standard VIII

Physics

Snell's Law of Refraction

Light moves f...

Question

Light moves from air into water. The light makes an angle of $47^{\circ}$ with the normal in the air. What angle will it make with the normal in the water?
The index of refraction for air is $1.00$ .
The index of refraction for water is $1.33$

47^{\circ}

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77^{\circ}

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43^{\circ}

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25^{\circ}

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33^{\circ}

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Solution

The correct option is E $33^{\circ}$
Simply from Snell's law we have,
$μ_{a i r} s i n i_{a i r} = μ_{w a t e r} s i n i_{w a t e r}$
$⟹ 1.00 \times s i n 47^{\circ} = 1.33 \times s i n i_{w a t e r}$
$⟹ i_{w a t e r} = 33^{\circ}$

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Light moves from air into water. The light makes an angle of 47∘ with the normal in the air. What angle will it make with the normal in the water?The index of refraction for air is 1.00.The index of refraction for water is 1.33

The correct option is E 33∘Simply from Snell's law we have,μairsiniair=μwatersiniwater⟹1.00×sin47∘=1.33×siniwater⟹iwater=33∘

Light moves from air into water. The light makes an angle of $47^{\circ}$ with the normal in the air. What angle will it make with the normal in the water?
The index of refraction for air is $1.00$ .
The index of refraction for water is $1.33$

The correct option is E $33^{\circ}$
Simply from Snell's law we have,
$μ_{a i r} s i n i_{a i r} = μ_{w a t e r} s i n i_{w a t e r}$
$⟹ 1.00 \times s i n 47^{\circ} = 1.33 \times s i n i_{w a t e r}$
$⟹ i_{w a t e r} = 33^{\circ}$