Case study (4 Marks)

Question 14 Marks

Draw the graph of deviation angle $(\delta)$ versus incidence angle $(i)$ and derive the formula for refractive index $n_{21}=\frac{\sin \left(\frac{A+D_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$ for the material of prism.

Answer

$\rightarrow$ The graph of deviation angle versus incidence angle is shown in figure.
$\rightarrow$ The graph shows that for a single value of deviation angle $(\delta)$ there are two values of incidence angle $i$ and hence also of $e$.
$\rightarrow$ From the symmetry it can be said that angle of deviation $\delta$ remains the same if angle of incidence $i$ and angle of emergent $e$ are interchanged.
Even if the path of ray can be traced back, resulting in the same angle of deviation.
$\rightarrow$ From the graph, for a particular value of $i=e$ the angle of incidence, a single value of deviation is obtained.
At the minimum deviation, $D _m,$ the refracted ray becomes parallel to its base.
$\rightarrow$ So when $\delta= D _m$ and $i=e$, then $r_1=r_2$.
$\rightarrow$ For prism, $ A =r_1+r_2$
$\therefore A =2 r_1$
$\therefore r_1 =\frac{ A }{2}......(1)$
$\rightarrow$ and from $\delta=i+e- A$,
$ D _m=2 i- A$
$2 i= D _m+ A$
$i=\frac{ A + D _m}{2}......(2)$
$\rightarrow$ Applying Snell's law at incident point $Q ,$
$n_1 \sin i=n_2 \sin r_1$
$\rightarrow$ Substituting value of $r_1$ and $i$ from equation $(1)$ and $(2),$
$\therefore n_1 \sin \left(\frac{ A + D _m}{2}\right)=n_2 \sin \left(\frac{ A }{2}\right)$
$\therefore \frac{n_2}{n_1}=\frac{\sin \left(\frac{ A + D _m}{2}\right)}{\sin \frac{ A }{2}}$
$\therefore n_{21}=\frac{\sin \left(\frac{ A + D _m}{2}\right)}{\sin \frac{ A }{2}}$
$\rightarrow$ which is the formula to find the refractive index of the material of the prism.

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Question 24 Marks

Explain the refraction by an equilateral prism and derive the formula $\delta=i+e- A$ in case of triangular prism of glass.

Answer

$\rightarrow $ Figure shows the cross section of a prism.
$\rightarrow $ The path of a light passing through this prism is $\text{PQRS .}$
$\rightarrow $ The angle of incidence is $i$ and the angle of refraction is $r$ at the first side $AB .$
$\rightarrow $ The angle incidence is $r_2$ and the angle of emergence $($angle of refraction$)$ is $e$.
$\rightarrow $ Angle between the direction of emergent ray $RS$ and incident ray $PQ$ is called angle of deviation $(\delta)$.
$\rightarrow $ In $\square \text{AQNR}\ m \angle AQN = m \angle ARN =90^{\circ}$
$\rightarrow $ The sum of remaining two angles is $180^{\circ}$.
$\therefore \angle A +\angle QNR =180^{\circ}......(1)$
$\rightarrow $ For $\triangle QNR$,
$r_1+r_2+\angle QNR =180^{\circ}......(2)$
$\rightarrow $ Comparing equation $(1)$ and $(2),$
$\therefore \angle A +\angle QNR =r_1+r_2+\angle QNR$
$\therefore A =r_1+r_2......(3)$
$\rightarrow $ For $\triangle QMR , \delta$ is the exterior angle.
$\therefore \delta=\angle MQR +\angle MRQ......(4)$
but $ i=r_1+\angle MQR$
$\therefore \angle MQR =i-r_1$
and same way $\angle MRQ =e-r_2$.
$\rightarrow $ Substituting these two values in equation $(4),$
$\therefore \delta=i-r_1+e-r_2$
$\therefore \delta=i+e-\left(r_1+r_2\right)$
$\rightarrow $ From equation $(3),$
$\therefore \delta=i+e- A$

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Question 34 Marks

Explain how the image is formed by thin convex lens and derive $-\frac{1}{u}+\frac{1}{v}=\left(n_{21}-1\right)\left[\frac{1}{R_1}-\frac{1}{ R _2}\right] \#\#\# $ Derive lens maker's formula for thin convex lens.

Answer

$\rightarrow$ Figure $(a)$ shows the geometry of image formation by a convex lens.
$\rightarrow$ A point object $O$ is placed at a distance $u$ from the optical centre.
On the other side of the lens there is image $I$.
Here image distance is $v$. The radii of curvature of both surfaces of the lens are $R_1$ and $R_2$ respectively and the focal length of the lens is $f$.
$\rightarrow$ The image formation can be seen in terms of two steps :
$(i)$ The first refracting surface forms the image $I_1$ of the object $O. \ ($figure $b)$
$(ii)$ The image $I _1$ acts as a virtual object for the second surface. $($figure $c)$ that forms image at $I$ .
$\rightarrow$ For refraction at interface $\text{ABC} ,$
$\frac{n_1}{ OB }+\frac{n_2}{ BI _1}=\frac{n_2-n_1}{ BC _1}......(1)$
$\rightarrow$ A similar procedure applied to the interface $\text{ADC}$ gives,
$-\frac{n_2}{ DI _1}+\frac{n_1}{ DI }=\frac{n_2-n_1}{ DC _2}$
$\rightarrow$ For a thin lens,
$BI _1= DI _1$
$\therefore -\frac{n_2}{ BI _1}+\frac{n_1}{ DI }=\frac{n_2-n_1}{ DC _2}......(2)$
$\rightarrow$ Adding equations $(1)$ and $(2),$
$\frac{n_1}{ OB }+\frac{n_1}{ DI }=\frac{n_2-n_1}{ BC _1}+\frac{n_2-n_1}{ DC _2}......(3)$
$\rightarrow$ Suppose the object is at infinity
i.e. $OB \rightarrow \infty$ and $DI \rightarrow f \ ($focal length$)$
$\rightarrow$from equation $(3),$
${c}0+\frac{n_1}{f}=\frac{n_2-n_1}{ BC _1}+\frac{n_2-n_1}{ DC _2}$
$\therefore \frac{n_1}{f}=\left(n_2-n_1\right)\left(\frac{1}{ BC _1}+\frac{1}{ DC _2}\right)$
$\rightarrow$ Now substituting $BC _1= R _1$ and $DC _2=- R _2$ in above equation.
$($Positive and negative signs are determined according to the sign convention$).$
$\therefore \frac{n_1}{f}=\left(n_2-n_1\right)\left(\frac{1}{ R _1}-\frac{1}{ R _2}\right)$
$\therefore \frac{1}{f}=\left(\frac{n_2-n_1}{n_1}\right)\left(\frac{1}{ R _1}-\frac{1}{ R _2}\right)$
$\therefore \frac{1}{f}=\left(\frac{n_2}{n_1}-1\right)\left(\frac{1}{ R _1}-\frac{1}{ R _2}\right)$
$\therefore \frac{1}{f}=\left(n_{21}-1\right)\left(\frac{1}{ R _1}-\frac{1}{ R _2}\right)$
$\rightarrow$This equation is known as lensmaker's formula.
$\rightarrow$Note that the formula is true for a concave lens also.
For concave lens $R_1$ is negative, $R_2$ positive and therefore $f$ is negative.

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Question 44 Marks

Explain refraction at spherical surfaces and derive formula $-\frac{n_1}{n}+\frac{n_2}{v}=\frac{n_2-n_1}{ R }$.

Answer

$\rightarrow $ As shown in figure, a point like object $O$ is placed on the principal axis of the spherical surface.
A spherical surface has centre of curvature $^{\prime} C\ '$ and radius of curvature $R.$
$\rightarrow $ Rays emerge from a medium having refractive index $n_1$.
Here, $OM$ and $ON$ are the incident rays.
$\rightarrow $ They refract in a medium having refractive index $n_2$.
Here $NI$ and $MI$ are the refractive rays. As a result, image $I$ of the point object $O$ is obtained.
$\rightarrow $ Assume that the aperture of the spherical surface is small compared to the object distance, image distance and radius of curvature, so that the angles can be taken small.
$\rightarrow $ Since the aperture of the surface is assumed to be small here, $NM$ will be taken to be nearly equal to the length of the perpendicular from the point $N$ on the principal axis.
$\rightarrow $ From figure,
$\tan \angle NOM \approx \angle NOM =\frac{ MN }{ OM }......(1)$
$\tan \angle NCM \approx \angle NCM =\frac{ MN }{ MC }......(2)$
$\tan \angle NIM \approx \angle NIM =\frac{ MN }{ MI }......(3)$
$\rightarrow $ For $\triangle NOC , i$ is the exterior angle.
Therefore,
$i=\angle NOM +\angle NCM$
Substituting values from equation $(1)$ and equation $(2),$
$\therefore i=\frac{ MN }{ OM }+\frac{ MN }{ MC }......(4)$
$\rightarrow $ From figure for $\triangle NIC , \angle NCM$ is the exterior angle.
$\therefore \angle NCM =r+\angle NIM$
$ r=\angle NCM -\angle NIM$
$\therefore r=\frac{ MN }{ MC }-\frac{ MN }{ MI }......(5)$
$\rightarrow $ By applying Snell's law at point $N,$
$n_1 \sin i=n_2 \sin r$
But $, \sin i \approx i$
$\therefore \sin r \approx r$
$\therefore n_1 i=n_2 r$
$\rightarrow $ Substituting $i$ and $r$ from equation $(4)$ and equation $(5),$
$\therefore n_1\left(\frac{ MN }{ OM }+\frac{ MN }{ MC }\right)=n_2\left(\frac{ MN }{ MC }-\frac{ MN }{ MI }\right)$
$\therefore \frac{n_1}{ OM }+\frac{n_1}{ MC }=\frac{n_2}{ MC }-\frac{n_2}{ MI }$
$\therefore \frac{n_1}{ OM }+\frac{n_2}{ MI }=\frac{n_2}{ MC }-\frac{n_1}{ MC }$
$\therefore \frac{n_1}{ OM }+\frac{n_2}{ MI }=\frac{n_2-n_1}{ MC }$
$\rightarrow $ But from figure, applying Cartesian sign convention,
$ OM =-u, MI =v $ and $ MC = R$
$\therefore -\frac{n_1}{u}+\frac{n_2}{v}=\frac{n_2-n_1}{ R }$
$\rightarrow $ Above equation gives us a relation between object and image distance in terms of refractive index of the medium and the radius of curvature of the curved spherical surface.

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Physics Case study (4 Marks)

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