Refraction of Light

Refraction of light

**Prism**

A transparent refracting medium which is bounded by five plane surfaces and having a triangular cross section is known as prism.

The figure below shows the passage of light through a triangular prism ABC.

The angles of incidence and refraction at first face AB are ∠*i* and ∠*r*_{1}.

The angle of incidence at the second face AC is ∠*r*_{2} and the angle of emergence ∠*e*.

*δ* is the angle between the emergent ray RS and incident ray PQ and is called the angle of deviation.

Here, ∠PQN = *i*

∠RQO = *r*_{1}

∠QRO = *r*_{2}

∠KTS = *δ*

∠TQO = *i* and ∠RQO = *r*_{1}, we have

∠TQR = *i − **r*_{1}

∠TRO = *e* and ∠QRO = *r*_{2}

∠TRQ = *e *− *r*_{2}

In triangle TQR, the side QT has been produced outwards. Therefore, the exterior angle *δ *should be equal to the sum of the interior opposite angles.

i.e, δ= ∠TQR + ∠TRQ = (*i* − *r*_{1}) + (*e* − *r*_{2})

δ = (*i *+ *e*) − (*r*_{1}+ *r*_{2}) …(i)

In triangle QRO,

*r*_{1}+ *r*_{2}+ ∠ROQ = 180° …(ii)

From quadrilateral AROQ, we have the sum of angles (∠AQO + ∠ARO = 180°). This means that the sum of the remaining two angles should be 180°.

i.e , ∠*A* + ∠QOR = 180° [∠A is called the angle of prism]

From equations (i) and (ii),

*r*_{1}+ *r*_{2} = *A* (iii)

Substituting (iii) in (i), we obtain

*δ* = (*i* + *e*) − *A*

$A+\delta =i+e$

If the angle of incidence is increased gradually, then the angle of deviation first decreases, attains a minimum value (*δ*_{m}), and then again starts increasing.

When angle of deviation is minimum, the prism is said to be placed in the minimum deviation position.

There is only one angle of incidence for which the angle of deviation is minimum.

When

*δ* = *δ*_{m} [prism in minimum deviation position],

*e* = *i* and *r*_{2} = *r*_{1} = *r* …(iv)

$\because {r}_{1}+{r}_{2}=A$

From equation (iv), *r* + *r* = *A*

$r=\frac{A}{2}$

Also, we have

*A* + *δ* = *i* + *e*

Setting,

*δ* = *δ*_{m} and *e = i*

*A* + *δ*_{m}_{ }= *i + i*

$i=\frac{\left(A+{\delta}_{\mathrm{m}}\right)}{2}\phantom{\rule{0ex}{0ex}}\because \mathrm{\mu}=\frac{\mathrm{sin}i}{\mathrm{sin}r}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{\mu}=\frac{\mathrm{sin}\left({\displaystyle \frac{A+{\delta}_{\mathrm{m}}}{2}}\right)}{\mathrm{sin}\left({\displaystyle \frac{A}{2}}\right)}$

**Factors affecting angle of deviation **

- Angle of incidence (
*i*) - Angle of prism (
*A*) - Refractive index ($\mu $) of the material of the prism
- Colour or wavelength ($\lambda $) of light

**Prism**

A transparent refracting medium which is bounded by five plane surfaces and having a triangular cross section is known as prism.

The figure below shows the passage of light through a triangular prism ABC.

The angles of incidence and refraction at first face AB are ∠*i* and ∠*r*_{1}.

The angle of incidence at the second face AC is ∠*r*_{2} and the angle of emergence ∠*e*.

*δ* is the angle between the emergent ray RS and incident ray PQ and is called the angle of deviation.

Here, ∠PQN = *i*

∠RQO = *r*_{1}

∠QRO = *r*_{2}

∠KTS = *δ*

∠TQO = *i* and ∠RQO = *r*_{1}, we have

∠TQR = *i − **r*_{1}

∠TRO = *e* and ∠QRO = *r*_{2}

∠TRQ = *e *− *r*_{2}

In triangle TQR, the side QT has been produced outwards. Therefore, the exterior angle *δ *should be equal to the sum of the interior opposite angles.

i.e, δ= ∠TQR + ∠TRQ = (*i* − *r*_{1}) + (*e* − *r*_{2})

δ = (*i *+ *e*) − (*r*_{1}+ *r*_{2}) …(i)

In triangle QRO,

*r*_{1}+ *r*_{2}+ ∠ROQ = 180° …(ii)

From quadrilateral AROQ, we have the sum of angles (∠AQO + ∠ARO = 180°). This means that the sum of the remaining two angles should be 180°.

i.e , ∠*A* + ∠QOR = 180° [∠A is called the angle of prism]

From e…

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