5 Marks Questions

Question 15 Marks

Mention the defects of human vision and describe the method to remove them.

Answer

The eye, as any other organ can suffer from problems or defects. we will consider four kinds of defects of vision, viz., short-sightedness or myopia; long-sighedness or hypermetropia; presbyopia; and astigmatism.
Short sightedness or Myopiya : This defect is also known as nearsightedness. A short-sighted person cannot see objects placed beyond a certain distance from the eye. This distance is the short-sighted eye's far point. Nearsightedness can occur because (a) the eyeball is elongated, i.e., the distance between the lens and then retina is greater than normal, or (b) the maximum focal length of the lens (when the eye is relaxed) is less than usual.

The result is that a parallel beam of light froma distant object converges short of the retina [Figure (a)]. However, a divergent beam from any point closer than the far point converges on the retina [Figure (b)]. This defect can be corrected by using a concave lens of appropriate focal length. The lens should be such that parallel rays from a distant object appear to come from the far point of the eye after refraction from it [(c)]. Suppose the focal length of the lens to be used is $f$. Then $u$ $=-\infty$ and $v=-d$ (the distance of the far point O from the concave lens), and from the lens formula,
$\frac{1}{-d}-\frac{1}{\infty}=\frac{1}{f} \Rightarrow f=-d$ ...(i)
The negative sign of the focal length shows that the lens is concave. The focal length of the lens should be equal to the distance of the far point of the eye.
Long-sightedness or hypermetropia : The defect is also known as farsightedness or hypermetropia. The eye, in this case, cannot see objects within a certain distance from the eye. The minimum distance beyond which it can see clearly without straining is its near point. The possible causes of this defect are (a) the shortening of th eyeball, i.e., the distance between the lens and the retina is less than normal, or (2) the focal length of the lens is greater than usual.
Consequently, rays from any point, N', which is closer than the near point, N, converge behind the retina [Figure (a)], while rays from all points, N', beyond the near point converge on the retina [Figure (b)]. This defect can be corrected by using a convex lens of the appropriate focal length to make the divergent rays from an object placed at the normal near point, N, converge on the retina after refraction by the convex lens. Then rays from the object appear to come from the point N', the near point of the defective eye, and can be seen clearly [Figure (c)].

Suppose the focal length of the convex lens used is $f$. Then $u =- D$ and $v=-d$ the distance of the near point N' from the convex lens.) Thus, from the lens formula,
$\frac{1}{f}=\frac{1}{-d}-\frac{1}{- D }=\frac{d- D }{ D d} \Rightarrow f=\frac{ D d}{d- D }$ ...(ii)
In equation (ii), as d > D the focal length is positive, which shows that the lens used is convex. Since the image of an object placed at the normal near point is formed at the near point N' of the defective eye, the image of any object situated between N and N' wil be formed beyond the near point of the defective eye and seen clearly due to the eye's accomodating power.
Presbyopia : In old age, the understanding capacity of the eye lens decreases. Therefore, neither objects that are very close nor that are very far away are clearly visible. This defect of the eye is called presbyopia. This defect can be removed by using bifocal lenses.
Astigmatism : An eye with astimatism cannot see horizontal and vertical lines located at equal distances with equal clearity. This defect arises due to unequal curvatures of the vertical or horizontal parts of the cornea or retina. This defect is rectified by the use of cylindrical lenses.

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Question 25 Marks

Establish the formula for the equivalent focal length of the combination of two thin lenses placed in contact.

Answer

Equivalent Lens : The lens which makes the image of an object placed infront of it at the same distance as made by the combination of two lenses in contact is called equivalent lens.
Formula of equivalent focal length : If O be the object at a distance $u$ from the lens of focal length f₁ and its image is formed at $I^{\prime}$ by this lens at a distance $v^{\prime}$, we have
$\frac{1}{f_1}=\frac{1}{v^{\prime}}-\frac{1}{u}$ ...(1)

The image $I^{\prime}$ will now serve as a virtual object for the second lens of focal length f₂ and final image I will be formed as a distance v, hence for second lens.
$\frac{1}{f_2}=\frac{1}{v}-\frac{1}{v^{\prime}}$ ...(2)
Adding equations (1) and (2), we get
$\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{v}-\frac{1}{u}$ ...(3)
If F be the focal length of equivalent lens for which $u$ and $v$ are distance of object and image, then
$\frac{1}{F}=\frac{1}{v}-\frac{1}{u}$ ... (4)
Comparing equations (3) and (4), we get
$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$ ...(5)
This formula is also true when both lenses are concave or one is convex and other is concave.
From equation (5), we get
$\frac{1}{F}=\frac{f_1+f_2}{f_1 f_2} \Rightarrow F=\frac{f_1 f_2}{f_1+f_2}$ ... (6)
Equations (5) and (6) represent two different forms of the formula for the focal length of equivalent lens made by placing two lenses in contact.
Now the following different cases arises:
(1) When both Lenses are Convex : In this situation f₁ and f₂ both will be positive. Therefore, from equation (6),
$F =\frac{\left(+f_1\right)\left(+f_2\right)}{\left(+f_1\right)+\left(f_2\right)}$
$\Rightarrow \quad F =\left(\frac{ f _1 f _2}{ f _1+ f _2}\right)$ ...(7)
i.e. the sign of equivalent focal length F is also positive.
Hence in this situation equivalent lens will also behave as a convex lens.
(2) When both Lenses are Concave : In this situation f₁ and f₂ both will have negative sign. Therefore, from equation (6). we have
$F =\frac{\left(-f_1\right)\left(-f_2\right)}{\left(-f_1\right)+\left(-f_2\right)}$
$\Rightarrow \quad F=-\left(\frac{f_1 f_2}{f_1+f_2}\right)$ ...(8)
i.e., the sign of equivalent focal length F is also negative. Hence in this situation equivalent lens will also behave as a concave lens.
(3) When one Lens is Convex and the other is Concave : Let the lens L₁ be convex lens and L₂ a concave lens i.e. f₁ will be positive and f₂will be negative. Therefore, from equation (6), we have
$F =\frac{\left(f_1\right)\left(-f_2\right)}{\left(f_1\right)+\left(-f_2\right)}$
$\Rightarrow \quad F=\frac{f_1 f_2}{f_1-f_2}$ ... (9)
(a) If $f_1 < f_2$ i.e. focal length of convex lens is less than that of concave lens then F will be positive, hence the equivalent lens will behave as a convex lens.
(b) If $f_1 > f_2$ i.e. focal length of convex lens is greater than that of concave lens then F will negative, hence the equivalent lens will behave as a concave lens.

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Question 35 Marks

Explain with a neat diagram construction and working of a compound microscope. Obtain an expression for its magnifying power.

Answer

Compound Microscope : In order to obtain higher magnifying power compound microscope is used.
Construction : It consists of two converging lenses. The lenses themselves are a suitable combination of lenses which avoids the various aberrations. The lens which is towards the object is called objective lens and the lens which is towards the eye is called eye piece. The objective lens has a wide aperture so as to cover up a wide field of vision and the eye-piece is of small aperture. In this microscope two convex lenses are used and magnification takes place in two steps hence it is called compound microscope.

Both the lenses are fitted in brass tubes, one of the tubes can slide into the other. This distance of eye-piece from the objective can be changed by moving the tube containing eyepiece by rack and pinion arrangement. The tube containing the eye-piece is fitted with cross wires to make measurements. These are two fine wires at right angles to each other. The distance between the objective and eye-piece is so adjusted that the image formed by the objective falls on cross wires. The compound microscope used in Biology has different eye-pieces of different power and a plane mirror is used to illuminate the object by reflection.
Formation of Image : The image formation by compound microscope is shown in Fig. (a) given above. In this figure let AB be an object to be seen. It is placed infront of the objective at a distance slightly greater than the focal length of the objective lens. By the objective lens a real, inverted and magnified image A₁B₁ is formed. The distance of eye lens is so adjusted that image A₁B₁ lies within the focal length of the eye lens. The image A₁B₁acts as an object for eye lens and magnified virtual image A₂B₂ is formed by the eye-piece. Thus, the final image A₂B₂ is seen through the eye-piece which is formed at the least distance of distinct vision.
If the image A₁B₁ is formed at the focus F_e of the eye-piece then the final image will be form at infinity.
Magnifying power : Magnifying power of a microscope is given by :
$M =\frac{\begin{array}{c}\text { Anglesubtended bythe } \\ \text { finalimageattheeye }\end{array}}{\begin{array}{c}\text { Anglesubtendedby theobject } \\ \text { placed attheleastdistance of distinct } \\ \text { visionandisbeing vieweddirectly }\end{array}}$
Let final image A₂B₂ subtend an angle $\beta$ at the eye lens E. Since eye is very near to eye lens, hence this angle $\beta$ can also be taken as subtended at the eye. Let the angle subtended at eye be a when the object AB is at least distance of distinct vision D.
Thus $M=\frac{\beta}{\alpha}=\frac{\tan \beta}{\tan \alpha}$
(since angle $\beta$ and $\alpha$ are very small)
$=\frac{A_1 B_1 / E A_1}{A B / D}=\left(\frac{A_1 B_1}{A B}\right) \times\left(\frac{D}{E A_1}\right)$
If u₀ and v₀ be the distances of object and image from objective lens, then from magnification formula of lens (taking proper sign) $\frac{ A _1 B_1}{ AB }=\frac{+v_o}{-u_o}$. If u_e be the distance of A₁B₁ from eye lens then EA₁ = - u_e. Hence

$M =-\frac{v_o}{u_o}\left(\frac{- D }{-u_e}\right)$
$\Rightarrow \quad M =-\frac{v_o}{u_o}\left(\frac{ D }{u_e}\right)$ ...(1)
If final image A₂B₂ is formed at $v_e$ from the eye lens then for this lens :
$v=v_e ; u=-u_e$ and $f=+f_e$
where $f_e$ = focal length of the eye lens. Then lens formula :
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{-v_e}-\frac{1}{-u_e}=\frac{1}{f_e}$
$\Rightarrow \quad \frac{1}{u_e}=\frac{1}{v_e}+\frac{1}{f_e}=\frac{1}{v_e}\left(1+\frac{v_e}{f_e}\right)$
Now substituting this value of $1 / u _e$ in above equation (1), we get
$M =-\left(\frac{v_o}{u_o}\right) \times D \left[\frac{1}{v_e}\left(1+\frac{v_e}{f_e}\right)\right]$
$\Rightarrow \quad M =-\left(\frac{ v _o}{ u _o}\right)\left(\frac{ D }{ v _{ e }}\right)\left( 1 +\frac{ v _{ e }}{ f _{ e }}\right)$ ...(2)
Now the following two cases are possible :
(i) If final image A₂B₂is formed at least distance of distinct vision D then v_e = D and from equation (2),
$M =-\left[\frac{ v _o}{u_o}\left( 1 +\frac{ D }{f_e}\right)\right]$ ...(3)
(ii) For relaxed eye final image A₂B₂ is formed at infinity In this case image A₁B₁ will be at the focus of eye lens $E \Rightarrow u _e= f _e$. Substituting this value of u_e in equation (1), we get
$M =-\left[\frac{ v _o}{u_o}\left(\frac{ D }{ f _e}\right)\right]$ ...(4)
Image formation ray diagram in this case is shown in fig. (c).
The negative sign in the formulae (3) and (4) for magnifying power in both the cases represents that final image formed in both the cases is inverted with respect to the object.

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Question 45 Marks

Write the formula for refraction of light at spherical surface (concave or convex). With the help of this formula derive the lens formula : $\frac{1}{f}=(n-1)(\frac{1}{R_{1}}-\frac{1}{R_{2}})$, where symbols used have their usual meanings.

Answer

Formula for refraction at spherical surface (concave or convex) is as follows :
$\frac{n}{v}-\frac{1}{u}=\left(\frac{n-1}{ R }\right)$
where $n$ = refractive index of the medium behind the surface with respect to the medium infront of the surface, R = radius of the curvature of spherical surface, $u$ = distance of object from the surface and $v$ = distance of the image from the surface.
Derivation of the formula : $\frac{1}{f}=(n-1)\left(\frac{1}{ R _1}-\frac{1}{ R _2}\right)$
In the ray diagram given below a convex lens 'L' made of absolute refractive index $n_2$ is placed in rarer medium (say air) of absolute refractive index $n_1$ i.e. $n_1 < n_2$.

In this figure P₁ , P₂are the poles and C₁ , C₂ are the centres of curvature of the lens. Hence P₁C₁ = R₁ and P₂C₂ = R₂ are the radii of curvatures of first and second surface of the lens respectively marked as (1) and (2) and P₁P₂ = t is the thickness of the lens. Let C be the optical centre of the lens.
Suppose O is a point object placed on the principal axis of the lens at a distance u from the pole P₁ of surface (1) of the lens. A ray OA is incident on this surface. At A it is refracted from rarer medium into denser medium bending towards normal C₁AN₁ at A along AB inside the lens. If the surface (2) were absent the ray AB would have met the principal axis at $I^{\prime}$ at a distance $v^{\prime}$ from the pole P₁. Hence $I^{\prime}$ can be treated as the real image formed by surface (1). On the basis of formula for refraction at single refracting surface for refraction at surface (1) from rarer medium of refractive index n₁ to denser medium of refractive index n₂ we have
$\frac{n_2}{v^{\prime}}-\frac{n_1}{u}=\frac{n_2-n_1}{R_1}$ ...(1)
The ray AB travelling in medium n₂ is actually incident on the second surface of the lens and is refracted into the rarer medium n₁, bending away from the normal C₂BN₂ at the point B. The emergent ray meets the principal axis at I which is the final, real image of O formed by the lens.
For refraction at the second surface of the lens, $I^{\prime}$ acts as virtual object and I is the real image of $I^{\prime}$ formed by this surface. Let v be the distance of I from the pole P₂ of the second surface. The distance of $I^{\prime}$ (object for the second surface) from the pole P₂ of the second surface is $\left(v^{\prime}-t\right)$. Then, for refraction at the surface (2) from denser medium $n_2$ to rarer medium $n_1$, we have
$\frac{n_1}{v}-\frac{n_2}{v^{\prime}-t}=\frac{n_1-n_2}{ R _2}$
But for thin lens thickness t << $v^{\prime}$ , hence this can be neglected, so we have
$\frac{n_1}{v}-\frac{n_2}{v^{\prime}}=\left(\frac{n_1-n_2}{ R _2}\right)$
or $\frac{n_1}{v}-\frac{n_2}{v^{\prime}}=-\left(\frac{n_2-n_1}{R_2}\right)$ ...(2)
On adding equation (1) and (2) we get
$\frac{n_1}{v}-\frac{n_1}{u}=\left(n_2-n_1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Dividing both sides by n_{1 ,}we get
$\frac{1}{v}-\frac{1}{u}=\left(\frac{n_2}{n_1}-1\right)\left(\frac{1}{ R _1}-\frac{1}{ R _2}\right)$
But n₁/n₂= n the refractive index of the material of the lens with respect to the surrounding medium,
$\therefore \quad \frac{1}{v}-\frac{1}{u}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ... (3)
But when the object is at infinity the image will be formed at the second focus of the lens i.e., when $u =\infty$, $v$ $=f$ the focal length of the lens. Putting these values of u and v in equation (3), we get
$\frac{1}{f}-\frac{1}{\infty}=(n-1)\left(\frac{1}{ R _1}-\frac{1}{ R _2}\right)$
$\Rightarrow \quad \frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ ...(4)
Equation (4) represents the lens maker's formula.
It has been derived for a convex lens forming a real image, but it is equally applicable to a convex lens forming a virtual image and to a concave lens which forms only virtual image.

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Question 55 Marks

Draw a graph between angle of deviation '$\delta$' and angle of incidence i for a prism showing the variation of $\delta$ with i.
Show for a prism that the refractive index n is given by :
$n=\frac{\sin \left[\left(\frac{A+\delta_m}{2}\right)\right]}{\sin (A / 2)}$
where symbols have their usual meanings.

Answer

Angle of deviation $\delta=i+e- A$. In this relation for a prism angle of prism A remains constant. Hence angle of deviation depends only on angle of incidence i because angle of emergence 'e' also depends upon i.
Now starting with the angle of incidence not less than 30° and increasing it by 5° interval the corresponding values of angle of deviation are determined. Now a graph is plotted between different values of angle of incidence i and corresponding value of angle of deviation $\delta$ which is the curve known as $(i-\delta)$ curve as shown in adjacent figure (a)

It is seen from the graph that as angle of incidence increases the angle of deviation is found to decreases in the beginning until it reaches the maximum value. The minimum value of angle of deviation is called the angle of minimum deviation and is denoted by $\delta_m$. Angle of deviation further increases with increase in angle of incidence.
Derivation of formula for refractive index 'n' of the material of the prism : In fig. (b) refraction of a light ray through a prism of refracting angle A is shown in which ABC represents the principal section of the prism. Here PQ is the incident ray falling on surface AB at the point Q making an angle of incidence ∠PQN = i with the normal NQE at Q to this surface AB. The ray PQ is refracted through prism along QR such that angle of refraction at Q is ∠EQR = r₁. The refracted ray QR meets the other surface AC at point R where it makes an angle ∠QRE = r₂as the angle of incidence with the normal MRE at point R of this surface AC. Due to refraction at this surface AC finally the ray emerges out of the prism along RS making an angle ∠MRS = e with the normal as angle of refraction and this angle is called as angle of emergence and RS as emergent ray. Thus PQRS is the complete path of the ray. By producing the incident ray PQ straight forward and the emergent ray RS straight backward they meet at the point O and at this point ∠DOR = $\delta$. This angle is called angle of deviation.

"Thus, angle of deviation is defined as the angle between the direction of incident ray and the direction of emergent ray."
Expression for angle of deviation: Since angle $\angle DOR$ $=\delta$ is the exterior angle of $\Delta OQR$ , hence it will be equal to the sum of two opposite interior angles.
$\therefore \quad \delta=\angle OQR +\angle ORQ$
$=(\angle OQE -\angle EQR )+(\angle ORE -\angle QRE )$
$=\left(\angle PQN { }_1-\angle EQR \right)+(\angle MRS -\angle QRE )$
( ∵ vertically opposite angles are equal)
$\Rightarrow \quad \delta=\left(i-r_1\right)+\left(e-r_2\right)$
$\Rightarrow \quad \delta=(i+e)-\left(r_1+r_2\right)$ ...(1)
But in the quadrilateral AQER the angles $\angle AQE$ and $\angle ARE$ are right angles, hence the sum of its remaining two angles will be $180^{\circ}$, i.e., $\angle A +\angle E =180^{\circ}$ ...(2)
But in $\triangle Q E R$, we have
$\angle RQE +\angle QRE +\angle QE R =180^{\circ}$
i.e. $\angle r_1+\angle r_2+\angle E =180^{\circ}$ ...(3)
Thus, comparing the equations (2) and (3), we get
$\angle r_1+\angle r_2=\angle A$ ...(4)
Now, on substituting this value of $\left(r_1+r_2\right)$ in equation (1) , we get
$\delta=( i + e )- A$ ...(5)

But the prism produces minimum deviation for one and only one particular value of angle of incidence. From the graph given in Fig. (a), it is obvious that for any angle of deviation the angle of incidence has two values as i and e. However, in the minimum deviation position i = e. In other words we can say that the deviation is minimum when the incident ray and the emergent ray are symmetrical with respect to their corresponding refracting surfaces. Hence in the position of minimum deviation the angles r₁ and r₂ become equal say equal to r and thus, the refracted ray inside the prism becomes parallel to the base of the prism [Fig. (c)].
In the minimum deviation position of the prism as discussed above the angle of emergence is equal to the angle of incidence, i.e., $e=\dot{r} ; r_1=r_2=r$ (say) and $\delta=\delta_m$.
Hence from equation (4),
$r_1+r_2= A \Rightarrow r+r= A$
$\Rightarrow \quad 2 r= A$ i.e. $r = A / 2$
and from equation (5),
$\delta=(i+e)- A$
$\Rightarrow \quad \delta_m=(i+i)- A$
i.e. $i=\left( A +\delta_m\right) / 2$
But according to Snell's law the refractive index 'n' of the material of the prism is given by :
$n=\frac{\sin i}{\sin r}$
Now, substituting the above values of $i$ and $r$ in this formula, we get
$n=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}$ ...(6)
(This is the required formula.)

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Question 65 Marks

What is dispersion of light? Explain its cause. Deviation of red colour is minimum white that for violet colour is minimum due to dispersion by prism. Explain it. Draw a diagram for dispersion.

Answer

Dispersion of Light and its cause : In the year 1666 Sir Issac Newton allowed the white light from sun to enter a darkened room through a small aperture S in a window and placed a glass prism in the path of the light rays. The light coming out of the prism was received on a white screen. A series of coloured bands was seen on the screen. The order of the colours from the side of base of the prism was as follows:
Violet, Indigo, Blue, Green, Yellow, Orange and Red. This order of colours in the spectrum can easily be remembered by the word VIBGYOR, the first letter of each colour.
This series of coloured bands is termed as spectrum. In the spectrum each colour is overlapped by the other i.e., there is no separate boundary line of a particular colour. But in fig. showing dispersion of white light through prism, colours are shown separated just for clarity for a single incident white light ray coming from the sun.

Dispersion of light may be defined as follows :
"The phenomenon of splitting of white light into its constituent colours on passing through a transparent medium i.e., prism is called dispersion of light and the series of coloured bands received on the screen is called spectrum of white light."
Cause of Dispersion : The wave theory of light has been able to explain satisfactorily the cause of phenomenon of dispersion. The different colours (i.e. constituents) of white light are in fact, the waves of different wavelengths. The wavelength of the visible region of light range between about 4000 Å for violet to 7800 Å for red. Even though the velocity of light is the same for all colours i.e., wavelengths in the vacuum but it is different for different colours in a transparent material medium. Now, on the basis of wave theory the refractive index n of a medium is defined as follows :
$n=\frac{\text { Velocityof lightinvacuum }}{\text { Velocity of lightinmedium }}$
i.e. $n=\frac{c}{V}$
As the velocity of light for a larger wavelength is found to be greater than that for a smaller wavelength i.e., $\lambda_{ V }<\lambda_{ R }$ , the refractive index for the larger wavelength is smaller than for the smaller wavelength i.e., $n_{ V }>n_{ R }$ , where V and R refer to violet and red colours.
Now, for thin prism, angle of deviation is given by :
$\delta=(n-1) A$
Since refractive index n is different for different colours, hence angle of deviation $\delta$ will be different for different colours. Hence if white light passes through the prism the rays of different colours are deviated through different angles. This causes splitting of light into its constituent colours. This is dispersion of white light.
Since refractive index of the material of the prism n is maximum for violet colour and minimum for red colour (i.e., $n_{ V }>n_{ R }$) hence deviation is maximum for red colour while it is minimum for red colour. (i.e., $\delta_{ V }>\delta_{ R }$).

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Question 75 Marks

An object is placed infront of a concave mirror of radius of curvature 15 cm at a distance of (i) 10 cm and (ii) 5 cm from the mirror. Calculate the position, nature and magnification of the image in each case.

Answer

Here $R =-15 cm \Rightarrow f= B / 2=-15 cm / 2$ $=-7.5 cm$
(i) When $u=-10 cm$
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \quad \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-7.5 cm}-\frac{1}{-10 cm}$
$=\left(\frac{-1}{30}\right) cm ^{-1} \Rightarrow v=-30 cm$
∴ Magnification :
$m=-\left(\frac{v}{u}\right)=-\left[\frac{-30 cm}{-10 cm}\right]=-3$
i.e. image will be formed at a distance of 30 cm for the mirror infront of it and will be real and inverted and three times in size in comparison to the size of the object.
(ii) When u = - 5 cm :
$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\left(\frac{1}{-7.5 cm}-\frac{1}{-5 cm}\right)=\left(\frac{1}{15}\right) cm ^{-1}$
$\Rightarrow \quad v=15 cm$
Magnification. $m=-\left(\frac{v}{u}\right)=\left[\frac{+15 cm}{-5 cm}\right]=+3$
i.e. image will be formed at a distance of 15 cm from the mirror behind it and it will be virtual, erect and three times in size in comparison to the size of the object.

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Question 85 Marks

Draw a ray diagram of image formation by a concave mirror and establish the relation between object distance $(u)$, image distance ($v$) and focal length $(f)$.

Answer

In adjacent figure, M₁M₂ is a concave mirror of small aperture whose pole is P and centre of curvature is C. An object AB is placed on its principal axis, beyond C i.e. between infinity and C. Here a ray AD from the object is drawn parallel to principal axis of the mirror, which after reflection at point D on the mirror passes through its focus F. Another ray AE is taken passing through the point C which strikes at the point E on the mirror normally, hence it is reflected back. The two reflected rays intersect each other at point A'. Thus A' is the real image of A. Now a perpendicular A'B' is drawn from A' to the principal axis. The image of any point on AB will lie on corresponding point on A'B'. Hence A'B' is the real image of AB formed by the reflection at concave mirror, which is inverted and smaller than AB.

Now, according to Cartesian sign convention, we have
Distance of object AB from P is PB = - u
Distance of image A'B' from P is PB' = - v
Focal length of the mirror PF = - f
Radius of curvature of the mirror PC = - R
Here AD || BP and since the distance between two parallel lines remains constant, hence AB = DN
Now $\triangle ABC$ and $\triangle A ^{\prime} B ^{\prime} C$ are similar triangles,
$\therefore \quad \frac{ AB }{ A ^{\prime} B ^{\prime}}=\frac{ CB }{ CB ^{\prime}}$ ...(1)
$\Delta DNF$ and $\Delta FA ^{\prime} B ^{\prime}$ are similar triangles,
$\therefore \quad \frac{ DN }{ A ^{\prime} B ^{\prime}}=\frac{ FN }{ FB ^{\prime}}$
But $AB=DN \text {, hence } \frac{AB}{A^{\prime} B^{\prime}}=\frac{FN}{FB^{\prime}}$ ... (2)
From equations (1) and (2), we have
$\frac{ CB }{ CB ^{\prime}}=\frac{ FN }{ FB ^{\prime}}$
Since aperture of the mirror is small, hence 'N' will be very close to F. Therefore,
$FN \approx FP$
$\therefore \quad \frac{ CB }{ CB ^{\prime}}=\frac{ FP }{ FB ^{\prime}} \quad \Rightarrow \quad \frac{ CB }{ CB ^{\prime}}=\frac{ PF }{ B ^{\prime} F }$
or $\frac{ PB - PC }{ PC - PB ^{\prime}}=\frac{ PF }{ PB ^{\prime}- PF } \Rightarrow \frac{-u-(- R )}{- R -(-v)}=\frac{-f}{-v-(-f)}$
$\Rightarrow \quad \frac{-u+ R }{- R +v}=\frac{-f}{-v+f} \Rightarrow \frac{-u+2 f}{-2 f+v}=\frac{-f}{-v+f}$
$(\because R=2 f)$
By cross multiplication, we have
$2 f^2-v f=u v-u f-2 v f+2 f^2$
$\Rightarrow 2 v f-v f+u f=u v$
$\Rightarrow \quad v f+u f=u v$
On dividing both sides by uvf, we get
$\frac{v f}{u v f}+\frac{u f}{u v f}=\frac{u v}{u v f} \Rightarrow \frac{ 1 }{ u }+\frac{ 1 }{ v }=\frac{ 1 }{ f }$
This is the required relation between $u,\ v$ and $f$ obtained for concave mirror.

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