4 Marks Questions

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

→The graph of deviation angle versus incidence angle is shown in figure.
→The graph shows that for a single value of deviation angle ( $\delta$ ) there are two values of incidence angle $i$ and hence also of $e$.
→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.
→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.
→So when $\delta= D _m$ and $i=e$, then $r_1=r_2$.
→For prism, $\begin{aligned} A & =r_1+r_2 \\ \therefore \quad A & =2 r_1 \\ \therefore \quad r_1 & =\frac{ A }{2}......(1)\end{aligned}$
→and from $\delta=i+e- A$,
$\begin{array}{l} D _m=2 i- A \\2 i= D _m+ A \\i=\frac{ A + D _m}{2}......(2)\end{array}$
→Applying Snell's law at incident point Q ,
$n_1 \sin i=n_2 \sin r_1$
→Substituting value of $r_1$ and $i$ from equation (1) and (2),
$\begin{array}{l}\therefore n_1 \sin \left(\frac{ A + D _m}{2}\right)=n_2 \sin \left(\frac{ A }{2}\right) \\\therefore \quad \frac{n_2}{n_1}=\frac{\sin \left(\frac{ A + D _m}{2}\right)}{\sin \frac{ A }{2}} \\\therefore \quad n_{21}=\frac{\sin \left(\frac{ A + D _m}{2}\right)}{\sin \frac{ A }{2}}\end{array}$
→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

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

→Figure (a) shows the geometry of image formation by a convex lens.
→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 $f$.
→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 .
→For refraction at interface ABC ,
$\frac{n_1}{ OB }+\frac{n_2}{ BI _1}=\frac{n_2-n_1}{ BC _1}......(1)$
→A similar procedure applied to the interface ADC gives,
$-\frac{n_2}{ DI _1}+\frac{n_1}{ DI }=\frac{n_2-n_1}{ DC _2}$
→For a thin lens,
$\begin{aligned}
& BI _1= DI _1 \\\therefore & -\frac{n_2}{ BI _1}+\frac{n_1}{ DI }=\frac{n_2-n_1}{ DC _2}......(2)\end{aligned}$
→Adding equations (1) and (2),
$\frac{n_1}{ OB }+\frac{n_1}{ DI }=\frac{n_2-n_1}{ BC _1}+\frac{n_2-n_1}{ DC _2}......(3)$
→Suppose the object is at infinity
i.e. $OB \rightarrow \infty$ and $DI \rightarrow f$ (focal length)
→from equation (3),
$\begin{array}{c}0+\frac{n_1}{f}=\frac{n_2-n_1}{ BC _1}+\frac{n_2-n_1}{ DC _2} \\\therefore \quad \frac{n_1}{f}=\left(n_2-n_1\right)\left(\frac{1}{ BC _1}+\frac{1}{ DC _2}\right)\end{array}$
→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). $\begin{array}{l}\therefore \quad \frac{n_1}{f}=\left(n_2-n_1\right)\left(\frac{1}{ R _1}-\frac{1}{ R _2}\right) \\\therefore \quad \frac{1}{f}=\left(\frac{n_2-n_1}{n_1}\right)\left(\frac{1}{ R _1}-\frac{1}{ R _2}\right) \\\therefore \quad \frac{1}{f}=\left(\frac{n_2}{n_1}-1\right)\left(\frac{1}{ R _1}-\frac{1}{ R _2}\right) \\ \therefore \quad \frac{1}{f}=\left(n_{21}-1\right)\left(\frac{1}{ R _1}-\frac{1}{ R _2}\right)\end{array}$
→This equation is known as lensmaker's formula.
→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

→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.
→Rays emerge from a medium having refractive index $n_1$. Here, OM and ON are the incident rays.
→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.
→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.
→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.
→From figure,
$\begin{array}{l}\tan \angle NOM \approx \angle NOM =\frac{ MN }{ OM }......(1) \\\tan \angle NCM \approx \angle NCM =\frac{ MN }{ MC }......(2) \\\tan \angle NIM \approx \angle NIM =\frac{ MN }{ MI }......(3)\end{array}$
→For $\triangle NOC , i$ is the exterior angle.
Therefore,
$i=\angle NOM +\angle NCM$
Substituting values from equation (1) and equation (2),
$\therefore \quad i=\frac{ MN }{ OM }+\frac{ MN }{ MC }......(4)$
→From figure for $\triangle NIC , \angle NCM$ is the exterior angle.
$\begin{aligned}\therefore & \angle NCM =r+\angle NIM \\& r=\angle NCM -\angle NIM \\\therefore \quad & r=\frac{ MN }{ MC }-\frac{ MN }{ MI }......(5)\end{aligned}$
→By applying Snell's law at point N,
$\begin{array}{c}n_1 \sin i=n_2 \sin r \\\text { But, } \sin i \approx i \\\therefore \quad \sin r \approx r \\\therefore \quad n_1 i=n_2 r\end{array}$
→Substituting $i$ and $r$ from equation (4) and equation (5),
$\begin{array}{l}\therefore n_1\left(\frac{ MN }{ OM }+\frac{ MN }{ MC }\right)=n_2\left(\frac{ MN }{ MC }-\frac{ MN }{ MI }\right) \\\therefore \quad \frac{n_1}{ OM }+\frac{n_1}{ MC }=\frac{n_2}{ MC }-\frac{n_2}{ MI } \\\therefore \quad \frac{n_1}{ OM }+\frac{n_2}{ MI }=\frac{n_2}{ MC }-\frac{n_1}{ MC } \\\therefore \quad \frac{n_1}{ OM }+\frac{n_2}{ MI }=\frac{n_2-n_1}{ MC }\end{array}$
→But from figure, applying Cartesian sign convention,
$\begin{aligned}& OM =-u, MI =v \text { and } MC = R \\\therefore \quad & -\frac{n_1}{u}+\frac{n_2}{v}=\frac{n_2-n_1}{ R }\end{aligned}$
→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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