Solution:

Ans. (i)

From fig $\mathrm{\angle }A+\mathrm{\angle }QNR={180}^{\circ }$

From triangle $\mathrm{△}\mathrm{QNR},r_1+r_2+\mathrm{QNR}={180}^{\circ }$

Hence from eqn. (i) & (ii)

$\therefore \mathrm{\angle }A=r_1+r_2$

The angle of deviation

$$\begin{array}{rlr}\delta & =(i-r_1)+(e-r_2)& =i+e-A\end{array}$$

At minimum deviation $i=e$ and $r_1=r_2$

$$\begin{array}{lll}\therefore & r=\frac{A}{2}\text{ And }& i=\frac{A+{\delta }_m}{2}\end{array}$$

Hence, refractive index,

$$\mu =\frac{\sin i}{\sin r}=\frac{\sin \left(\frac{A+{\delta }_m}{2}\right)}{\sin \frac{A}{2}}$$

(ii) From snell’s law, ${\mu }_1\sin i={\mu }_2\sin r$ Given

$${\mu }_1=\sqrt{2},{\mu }_2=1$$

and

$$\begin{array}{rlrl}r& ={90}^{\circ }\text{ (just grazing) }i& =\frac{1}{\sqrt{2}}i& ={45}^{\circ }\end{array}$$

$\Rightarrow \sin i=\frac{1}{\sqrt{2}}$

or

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AII Q. 4. (i) Plot a graph to show variation of the angle of deviation as a function of angle of incidence for light passing through a prism. Derive an expression for refractive index of the prism in terms of angle of minimum deviation and angle of prism.

(ii) What is dispersion of light? What is its cause ?

(iii) A ray of light incident normally on one face of a right isosceles prism is totally reflected as shown in fig. What must be the minimum value of refractive index of glass? Give relevant calculations.

R [Delhi I, II, III 2016]

Ans. (i)

From figure $\delta =D_{m^{\prime }}i=e$ which implies $r_1=r_2$

Using

$$ \begin{aligned} 2 r & =A, \text { or } r=\frac{A}{2} \ \delta & =i+e-A \ D_{m} & =2 i-A \ i & =\frac{\mathrm{A}+\mathrm{D}{m}}{2} \ \mu & =\frac{\sin i}{\sin r}=\frac{\sin \left(\frac{\mathrm{A}+\mathrm{D}{m}}{2}\right)}{\sin \frac{\mathrm{A}}{2}} 1 / 2 \end{aligned} $$

(ii) The phenomenon of splitting of white light into its constituent colours.

Cause: Refractive index of the material is different for different colours. According to the equation, $\delta$ (deviation) $=(\mu -1)A$, where, $A$ is the angle of prism, different colours will deviate through different amount. (iii)

For total internal reflection, $\mathrm{\angle }i\ge \mathrm{\angle }{i}_c$ (critical angle) $1/2$

$$\begin{array}{rlrlrl}\Rightarrow {45}^{\circ }& \ge \mathrm{\angle }{i}_{c^{\prime }}\text{ i.e., }\mathrm{\angle }{i}_c\le {45}^{\circ }1/2\sin i_c& \le \sin {45}^{\circ }& \le \frac{1}{\sqrt{2}}\frac{1}{\sin i_c}& \ge \sqrt{2}\Rightarrow \mu & \ge \sqrt{2}\end{array}$$

Hence, the minimum value of refractive index must be $\sqrt{2}$.

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TOPIC-3

Revision Notes

Based upon phenomenon of reflecting and refracting properties of mirrors, lenses and prisms, a number of optical devices and instruments have been designed.

Microscope is an optical instrument which help us to see and study micro objects or organism. It forms magnified image of the object.

Telescope is an optical instrument which help usta seeland study far off objects magnified and resolved (with clarity).

We generally set these instruments at two different image vision positions

• Image at least distance of distinct vision : This is the least distance from eye where we able to see objects distinctly. For normal human eye this distance is $25\mathrm{cm}$ from our eye.
• Image at relaxed vision : Thisis the distance from eye where we able to see objects distinctly in relax vision (no strain to eye). For normalnuman eye this distance is infinity from our eye.
• Magnification at distinct vision is always greater than magnification at relaxed vision.

Simple Microscope : Convex lens behaves as simple microscope.

The magnifying power of the simple microscope

(i) For least distance of distinct vision $m=1+\frac{D}{f}$

where, $D$ is the least distance of distinct vision of the eye. And $f$ is focal length of the lens.

(ii) For relaxed eye

$$m=\frac{D}{f}$$

from above formulae, it is clear that for larger magnifying power, the focal length of the convex lens should be small.

Please note that angular magnification by optical instruments is the linear magnification by lenses only. It means magnification of an instrument means how many times it enlarges the image of object. So this is just as

$$m=\frac{h^{\prime }}{h}$$

where, $h$ is size of object (in one dimension) and $h^{\prime }$ is the size of image.

Compound Microscope : For much large magnification, compound microscope is used. It is a combination of two convex lenses hence the magnification of each lens is compounded.

• These two lenses are placed co-axially and the distance between them is adjustable.
• The lens towards the object is called objective and that towards the eye is called eyepiece.
• The final image formed by the compound microscope is magnified and inverted.
• Total magnification by compound lens

$$m=m_o\times m_e$$

where, $m_0$ is magnification by objective lens and $m_e$ is magnifeation by eyepiece.

• For least distance of distinct vision magnification by objectivelens is

$$m_0=\frac{v_0}{u_0}\Leftarrow \frac{L}{f_0}$$

where, $L$ is the distance between the second focal point of the objective and the first focal point of the eyepiece (focal length $f_e$ ). It is called the tube length of the compound microscope.

Eyepiece lens will act as simple microscope.

Magnification by eyepiece lens is

$$m_{\mathrm{e}}=1+\frac{D}{f_e}$$

Hence,

For Relaxed eye (normal adjustment)

For relaxed eye the magnification by objective lens remain same, the magnification by eyepiece will be $+\frac{D}{f_e}$

Hence, the total magnification of compound microscope in relaxed eye condition is

$$m=\frac{L}{f_0}\times \frac{D}{f_e}$$

P Properties of Compound Microscope

• For large magnification of a compound microscope, both $f_0$ and $f_e$ should be small.
• If the length of the microscope tube increases, then its magnifying power increases.
• Generally $f_0$ is much smaller. So that objective is placed very near to principal focus.
• The aperture of the eyepiece is generally small so that whole of the light may enter the eye.
• The aperture of the objective is also small so that the field of view may be restricted.

> Magnification by Telescope

• Telescope is an instrument to magnify and resolve far off objects.
• Far off objects make much smaller angle at our eye. Telescope makes that angle bigger without much intensity loss.
• To maximise the intensity aperture size of objective lens is quite large. It will focus a bright point size image at its focal plane.
• Now with eyepiece, we will form this point size image to final inverted magnified image. This type of telescope is known as astronomical telescope.
• For least distance of distinct vision

$$m=-\frac{f_o}{f_e}(1+\frac{f_e}{D})$$

$>$ For relaxed eye (normal adjustment)

$$m=\frac{\alpha }{\beta }=-\frac{f_o}{f_e}$$

Properties of astronomical telescope

• For larger magnifying power, $f_0$ should be large and $f_e$ should be small.
• The length of the tube of anastronomical telescope is $L=f_o+f_e$ for relaxed vision adjustment.
• When the length of the the of the telescope increases, $f_0$ increases and hence the magnifying power also increases.

> Limitations of refractive telescope

• Large objectivelens makes the telescope very heavy. So it is difficult to handle it by hand.
• It has spherical and chromatic aberrations.

Modern Telescope (Reflective Telescope)

• Reflecting telescope consists of a concave mirror of large radius of curvature in place of objective lens
• A secondary convex mirror is used to focus the incident light, which now passes through a hole in the objective primary mirror .
• The magnifying power of the reflecting telescope is $m=\frac{f_0}{f_e}$

Advantages of reflective telescope

• Very sharp point image by objective mirror removes spherical aberrations.
• As it is very light so large aperture of parabolic mirror can be used for desired magnification.
• This is based on the principle of reflection hence there will be no chromatic aberrations.

Know the Formulae

Magnification by Simple Microscope

$m=1+\frac{D}{f}$ (for distinct vision)

$m=\frac{D}{f}$ (For relaxed eye )

• Magnification by Compound Microscope

$\frac{L}{f_0}(1+\frac{D}{f_e})$ or $\frac{v_o}{u_o}(1+\frac{D}{f_e})$ (for distinct vision )

$\frac{L}{f_0}\times \frac{D}{f_e}$ or $\frac{v_o}{u_o}\times \frac{D}{f_e}$ (for relaxed eye)

Magnification by Telescope

$m=-\frac{f_0}{f_e}(1+\frac{f_e}{D})$ (for distinct vision)

$m=-\frac{f_o}{f_e}$ (for relaxed eye)

Very Short Answer Type Questions

(1 mark each)

AI Q. 1. Does the magnifying power of a microscope depend on the colour of the light used ? Justify your answer.

U] [Foreign 2017]

Ans. Justification, $m\propto \frac{1}{f_o{f}_e}$

And focal length depends on colour $\mu$.

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Detailed Answer :

Magnifying power of microscope $\propto \frac{1}{f_0{f}_e}$ Where $f_0$ and $f_{\rho }$ are the focal lengths of objective lens and eyepiece lens respectively.

Focal length of a lens depends upon the refractive index and different colour of light has different refractive index with respect to the medium of lens material. Hence, magnifying power of a microscope depends on the colour of the light used.