Question

the angle of incidence for a ray of light from air to water interface is 40 degree. if the ray travels through the water with a refractive index of 1.33, calculate the angle of refraction.

Answer 1

`{ \qquad\qquad\huge\underline{{\sf Answer}}}`

Here's the solution ~

According to snells law :

`\qquad \sf  \dashrightarrow \: n_1 ×sin \: i = n_2 * sin \:r`

• i = angle of incidence
• r = angle of refraction

[ n corresponds to the refracting index of a medium ]

now, as we know :

• retractive index of air is approximately 1

• angle of incidence is given to be 40°

• refrctive endex of water is given 1.33

Now, let's proceed according to the formula ~

`\qquad \sf  \dashrightarrow \: 1 * \sin(40 \degree) = 1.33 * \sin(r)`

`\qquad \sf  \dashrightarrow \: \sin(r) = \cfrac{1 * sin \: r}{1.33}`

[ sin 40° = 0.6428 ]

`\qquad \sf  \dashrightarrow \: \sin(r) = \cfrac{0.6428}{1.33}`

`\qquad \sf  \dashrightarrow \: \sin(r) \approx \cfrac{1}{2}`

`\qquad \sf  \dashrightarrow \: r \approx 30 \degree`

Hence, the angle of refraction is equal to 30°