Snell's Law Calculator

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**Snell's Law** was named after the scientist **Willebrod Snellius** who first formulated. This law gives the idea about how the light behaves when it enters two different media.

When light travels from first medium to the second medium, it obeys the Snell's Law which is as below:

The law states that :

TheĀ**Snell's Law Formula** is given by

**Snell's Law Calculator** helps us to determine the unknown parameter among the above four parameters if any of the three parameters are given.

**Step 1 :**

**Step 2 :**

When light travels from first medium to the second medium, it obeys the Snell's Law which is as below:

The law states that :

- The incident ray, the refracted ray and the normal to the surface at the point of incidence all lie in one plane.
- For any two given pair of mediums, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant.

TheĀ

$\frac{sin_{\theta_{1}}}{sin_{\theta_{2}}}$ = $\frac{n_{2}}{n_{1}}$ = a Constant = $\mu$.

Where $\theta_{1}$ = Angle of Incidence in First Medium,

$\theta_{2}$ = Angle of Refraction in Second Medium,

$\mu$ = Constant,

n_{1} = Refractive Index in first medium,

n_{2} = Refractive Index in second medium.

$\theta_{2}$ = Angle of Refraction in Second Medium,

$\mu$ = Constant,

n

n

Go through the problem and analyze what are the given parameters and what is the Unknown quantity.

Use the formula

n_{1} sin $\theta_{1}$ = n_{2} sin $\theta_{2}$

substitute the given parameters in this formula and determine the Unknown quantity.

If light travels from Air to water with angle of incidence of 30

^{0}. Calculate its angle of refraction.**Step 1 :**Given that:

Refractive Index of Air n

_{1}=1,

Refractive Index of Water n_{2}= 1.33,

Angle of incidence $\theta_{1}$= 55^{0}**Step 2 :**Use the formula

n_{1}sin $\theta_{1}$ = n_{2}sin $\theta_{2}$

Substituting the Values

sin $\theta_{2}$ = $\frac{n_{1} sin \theta_{1}}{n_{2}}$

= $\frac{1 \times sin {30^{\circ}}}{1.33}$

**Answer :**0.727

A light ray enters in to the optical medium having R.I as 1.5 from air having an angle of Incidence of 40

^{0}. Calculte its angle of refraction.**Step 1 :**given $\theta_{1}$ = 40

^{0},

$\theta_{2}$ = ?,

n_{1}= 1,

n_{2}= 1.5**Step 2 :**sin $\theta_{2}$ = $\frac{n_{2} sin \theta_{2}}{sin \theta_{1}}$

= $\frac{1.5 \times sin 40^{\circ}}{1}$

sin $\theta_{2}$ = 0.4967**Answer :**$\theta_{2}$ = 0.5198