Q: Derive the Lens Maker's formula.

Consider the situation shown in figure. \$ADBE\$ is a thin lens. An object 0 is placed on its principal axis. The two spherical surfaces of the lens have their centres at \$C_1\$ and \$C_2\$. The optical centre is at \$P\$ and the principal axis cuts the two spherical surfaces at \$D\$ and \$E\$.As shown in the figure, after the first refraction image is formed at \$O_1\$ and after second refraction the final image is formed at \$I\$. Asthe lens is thin, the points \$ D, P and E\$ are all close to each other and we may take the origin at \$P\$ for both these refractions.The general equation for refraction at a spherical surface is\$\dfrac{\mu_2}{v}-\dfrac {\mu_1}{u} = \dfrac {\mu_2-\mu_1}{R}\$..........(i) For the first refraction, the object is at \$O\$, theimage is at \$O_1\$ and the centre of curvature is at \$C_1\$. If\$u\$, \$v\$, and \$R\$, denote their x-coordinates,\$\dfrac{\mu_2}{v_1}-\dfrac {\mu_1}{u} = \dfrac {\mu_2-\mu_1}{R_1}\$.......(ii)For the second refraction at \$AEB\$, the incident rays \$GH\$ and \$DE\$ diverge from \$O_1\$. Thus, \$ O_1 \$ is the object forthis refraction and its x-coordinate is \$v_1\$. The image isformed at \$I\$ and the centre of curvature is at \$C_2\$. Theirx-coordinates are \$v\$ and \$R\$ respectively. The light goesfrom the medium \$\mu_2\$ to medium \$\mu_1\$.Hence \$\dfrac{\mu_1}{v}-\dfrac {\mu_2}{v_2} = \dfrac {\mu_1-\mu_2}{R_2}\$.......(iii)Adding (ii) and (iii),\$\dfrac{1}{v}-\dfrac{1}{u}=(\dfrac{\mu_2}{\mu_1}-1)\times (\dfrac{1}{R_1}-\dfrac{1}{R_2})\$ If the object \$O\$ is taken far away from the lens, the image is formed close to the focus. Thus, for \$u\$= \$\infty\$ , \$v=f\$Hence,\$\dfrac{1}{f} = (\dfrac {\mu_2}{\mu_1}-1)\times(\dfrac{1}{R_1}-\dfrac{1}{R_2})\$ This is lens makers formula.

Q: A point object \$O\$ is placed in front of a glass rod having a spherical end of radius of curvature \$30\ cm\$. The image would be formed at :

\$\begin{array}{l}{\mu _1}({\rm{air}}) = 1\\{\mu _2}(glass) = 1.5\\u = - 15cm\\R = 30cm\\\dfrac{{{\mu _2} - {\mu _1}}}{R} = \dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u}\\\dfrac{{1.5 - 1}}{{30}} = \dfrac{{1.5}}{v} - \dfrac{1}{{15}}\\v = - 30\end{array}\$Ans. \$30cm\$ towards the left.

Q: A ray incident at a point at an angle of incidence of \${60}^{o}\$ enters a glass sphere of \$R.l.n=\sqrt 3\$ and is reflected and refracted at the further surface of the sphere. The angel between the reflected and refracted rays at this surface is

Refraction at \$P\$, \$ \dfrac { \sin { 60 }^{o} }{ \sin { { r }_{ 1 } } } =\sqrt { 3 } \$\$\Rightarrow \sin { { r }_{ 1 } } =1/2 \quad \Rightarrow { r }_{ 1 }=30\$Since \${ { r }_{ 2 } } ={ r }_{ 1 } \quad \therefore \quad{ r }_{ 2 }=30\$Reflection at \$Q\$, \$\dfrac { \sin { { r }_{ 2 } } }{ \sin { { i }_{ 2 } } } =\dfrac { 1 }{ \sqrt { 3 } } \$Putting \${R}_{2}={30}^{o}\$ we obtain \${i}_{2}={60}^{o}\$Reflection at \$Q, {r}^{\prime }_{2}={r}_{2}\$\$\therefore \quad \alpha =180^{o}-({ r }^{ \prime }_{ 2 }+{ r }_{ 2 })=180^{o}-(30^{o}+60^{o})=90^{o}\$Hence (C) is correct.

Question 1

Derive the Lens Maker's formula.

Accepted Answer

Consider the situation shown in figure. $ADBE$  is a thin lens. An object 0 is placed on its principal axis. The two spherical surfaces of the lens have their centres at $C_1$ and $C_2$. The optical centre is at $P$ and the principal axis cuts the two spherical surfaces at $D$ and $E$.As shown in the figure, after the first refraction image is formed at $O_1$ and after second refraction the final image is formed at $I$. Asthe lens is thin, the points $ D, P and E$  are all close to each other and we may take the origin at $P$ for both these refractions.The general equation for refraction at a spherical surface is$\dfrac{\mu_2}{v}-\dfrac {\mu_1}{u} = \dfrac {\mu_2-\mu_1}{R}$..........(i) For the first refraction, the object is at $O$, theimage is at $O_1$ and the centre of curvature is at $C_1$. If$u$, $v$, and $R$, denote their x-coordinates,$\dfrac{\mu_2}{v_1}-\dfrac {\mu_1}{u} = \dfrac {\mu_2-\mu_1}{R_1}$.......(ii)For the second refraction at $AEB$, the incident rays $GH$ and $DE$ diverge from $O_1$. Thus, $ O_1 $ is the object forthis refraction and its x-coordinate is $v_1$. The image isformed at $I$ and the centre of curvature is at $C_2$. Theirx-coordinates are $v$ and $R$ respectively. The light goesfrom the medium $\mu_2$  to medium $\mu_1$.Hence $\dfrac{\mu_1}{v}-\dfrac {\mu_2}{v_2} = \dfrac {\mu_1-\mu_2}{R_2}$.......(iii)Adding (ii) and (iii),$\dfrac{1}{v}-\dfrac{1}{u}=(\dfrac{\mu_2}{\mu_1}-1)\times (\dfrac{1}{R_1}-\dfrac{1}{R_2})$ If the object $O$ is taken far away from the lens, the image is formed close to the focus. Thus, for $u$= $\infty$ , $v=f$Hence,$\dfrac{1}{f} = (\dfrac {\mu_2}{\mu_1}-1)\times(\dfrac{1}{R_1}-\dfrac{1}{R_2})$ This is lens makers formula.

Question 2

A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such away that the end closer to the pole is 20 cm away from it. Find the length of the image.

Accepted Answer

The position of image of nearer end to mirror can be found out using mirror equation,$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$$\implies \dfrac{1}{v_1}=\dfrac{1}{-20}-\dfrac{1}{-10}$$\implies v_2=-20$Similarly position of image of farther end to mirror can be found out as$=\dfrac{1}{v_2}=\dfrac{1}{-(10+20)}-\dfrac{1}{-10}$$\implies v_2=-15$Hence the length of the image $=20cm-15cm=5cm$

Question 3

One half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Explain your observations.

Accepted Answer

When one-half of a convex lens is covered with a black paper, this lens produces a complete image of the object. To prove it we perform an experiment as given below: Take a convex lens and cover half part of it by using a black paper. Place it vertically in a stand. On one side of it place a burning candle. On opposite side of the lens fix a white screen. Adjust the position of candle or screen till clear image of burning candle is formed on the screen. We observe that the image is a complete image of the object (burning candle). From the experimental observations, we find that image formation does not depent upon the size of a lens. A smaller lens can also form complete image of an object placed in front of it. However, brightness of the image decreases when some part of lens is blocked. It is because now lesser number of rays pass through the lens.

Question 4

Derive Lens Maker's formula for a convex lens.

Accepted Answer

Formula for refraction of light from the spherical surface is $\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}$ where $OB=-u, OI=+v, BC=+R$$n_2$ is refractive index of medium 2 i.e. spherical glassand $n_1$ is refractive index of medium 1.Figure (a) shows the geometry of image formation by a double convex lens. the image formation can be seen in two steps, the first refracting surface form an image $I_1$ which act as a virtual object for the second surface. applying the equation above written for surface 1$\dfrac{n_1}{OB}+\dfrac{n_2}{BI_1}=\dfrac{n_2-n_1}{BC_2}$similarly, form the second interface $-\dfrac{n_2}{DI_1}+\dfrac{n_1}{DI}=\dfrac{n_2-n_1}{DC_2}$ (refractive index of right side medium is $n_1$ and for left side medium is $n_2$, and also $DI_1$is negative.)now for thin lens $BI_1=DI_1$. adding both the equations we get$\dfrac{n_1}{OB}+\dfrac{n_1}{DI}=(n_2-n_1)(\dfrac{1}{BC_1}+\dfrac{1}{DC_2})$and assuming the object at infinity means OB will be very large we get $\dfrac{n_1}{DI}=(n_2-n_1)(\dfrac{1}{BC_1}+\dfrac{1}{DC_2})$and dividing both sides with $n_1$$\dfrac{1}{DI}=(n_{21}-1)(\dfrac{1}{BC_1}+\dfrac{1}{DC_2})$replacing $ DI=f, BC_1=+R_1, DC_2=-R_2$Finally, we get $\dfrac{1}{f}=(n_{21}-1)(\dfrac{1}{R_1}+\dfrac{1}{R_2})$.This equation is known as lens makers formula.

Question 5

Draw a ray diagram to show the formation of image by a convex lens when an object is placed in front of the lens between its optical centre and principal focus.

Accepted Answer

When we place any object between the optic center and principal focal length the image is formed on the same side of the object and is virtual,erect and enlarged.

Question 6

Derive lens formula $\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ .

Accepted Answer

The figure above shows that the formation of a real, inverted and diminished image AB of the object AB placed beyond the centre of curvature at a distance u from the convex lens. Let v be the image distance,According to Cartesian sign conventionObject distance $(OB) = -u$Image distance $(OB) = +v$Focal length $(OF1 = OF2) = +f$From the geometry of the figure above, right angled $\triangle$ ABO and $\triangle$ ABO are similar.$\therefore \dfrac {AB} {A'B'} =  \dfrac {OB} {O'B'} = -v $ ........ (i)Similarly, from the geometry of the figure above, right angled $\triangle ODF_2$ and $\triangle BAF_2$ are similar.$\therefore \dfrac {OD} {A'B'} =  \dfrac {OF_2} {F_2B^2} $$\therefore \dfrac {AB} {A'B'} =  \dfrac {OF_2} {F_2B^2}$ ( $\because $ OD = AB, are the opposite sides of $\Box$ ABOD )$\therefore \dfrac {AB} {A'B'} =  \dfrac {OF_2} {OB' - OF_2} $$\therefore \dfrac {AB} {A'B'} =  \dfrac {f} {v- f} $ .......(ii)From (i) and (ii),$ - \dfrac {u} {v} = \dfrac {v} {v-f} $$\Rightarrow  -u (v - f) = vf$$\Rightarrow-uv + uf = vf$Dividing each term by $uvf,$$-\dfrac{1}{f} + \dfrac{1}{v} = \dfrac{1}{u}$$\dfrac{1}{v}-\dfrac{1}{u}= \dfrac{1}{f}$This equation is called the 'Lens formula'.

Question 7

A point object $O$ is placed in front of a glass rod having a spherical end of radius of curvature $30\ cm$. The image would be formed at :

Accepted Answer

$\begin{array}{l}{\mu _1}({\rm{air}}) = 1\{\mu _2}(glass) = 1.5\u =  - 15cm\R = 30cm\\dfrac{{{\mu _2} - {\mu _1}}}{R} = \dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u}\\dfrac{{1.5 - 1}}{{30}} = \dfrac{{1.5}}{v} - \dfrac{1}{{15}}\v =  - 30\end{array}$Ans. $30cm$ towards the left.

Question 8

A lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I$. Aperture of diameter $\dfrac {d}{2}$ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively.

Accepted Answer

By covering aperture, focal length does not change.But intensity is reduced by $\dfrac {1}{4}$ times, as aperture diameter $\dfrac {d}{2}$ is covered.$\therefore I' = I - \dfrac {I}{4} = \dfrac {3I}{4}$$\therefore$ New focal length $= f$ and intensity $= \dfrac {3I}{4}$.

Question 9

The distance between an object and the screen is $100\ cm$. A lens produces an image on the screen when the lens is placed at either of the positions $40\ cm$ apart. The power of the lens is nearly

Accepted Answer

At first position of lens, let the distance of lens from object and screen be $x$ and $y$, respectively.$\therefore x + y = 100 .... (i)$$\therefore$ At second position of lens, the distance of lens from object and screen shall be $y$ and $x$, respectively.$\therefore y - x = 40 .... (ii)$Solving Eqs. (i) and (ii), we get$y = 70\ cm = \dfrac {70}{100} m$ and $x = 30\ cm = \dfrac {30}{100} m$Therefore, the power of lens is$\dfrac {1}{f} = \dfrac {1}{y} + \dfrac {1}{x} = \dfrac {100}{70} + \left (\dfrac {100}{30}\right ) = \dfrac {100}{21} \approx 5\ diopter$.

Question 10

A ray incident at a point at an angle of incidence of ${60}^{o}$ enters a glass sphere of $R.l.n=\sqrt 3$ and is reflected and refracted at the further surface of the sphere. The angel between the reflected and refracted rays at this surface is

Accepted Answer

Refraction at $P$, $ \dfrac { \sin { 60 }^{o}  }{ \sin { { r }_{ 1 } }  } =\sqrt { 3 } $$\Rightarrow \sin { { r }_{ 1 } } =1/2 \quad \Rightarrow { r }_{ 1 }=30$Since ${ { r }_{ 2 } } ={ r }_{ 1 } \quad \therefore \quad{ r }_{ 2 }=30$Reflection at $Q$, $\dfrac { \sin { { r }_{ 2 } }  }{ \sin { { i }_{ 2 } }  } =\dfrac { 1 }{ \sqrt { 3 }  } $Putting ${R}_{2}={30}^{o}$ we obtain ${i}_{2}={60}^{o}$Reflection at $Q, {r}^{\prime }_{2}={r}_{2}$$\therefore \quad \alpha =180^{o}-({ r }^{ \prime  }_{ 2 }+{ r }_{ 2 })=180^{o}-(30^{o}+60^{o})=90^{o}$Hence (C) is correct.

Principle Rays For Lens

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REVISE WITH CONCEPTS

Convenient Rays to Draw Ray Diagrams

Image Formation by Lenses

QUICK SUMMARY WITH STORIES

Convenient Rays to Draw Ray Diagrams

Problem-based on Image Formation by Lens of Moving Object.

Derive the Lens Maker's formula.

A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such away that the end closer to the pole is 20 cm away from it. Find the length of the image.

One half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Explain your observations.

Derive Lens Maker's formula for a convex lens.

Draw a ray diagram to show the formation of image by a convex lens when an object is placed in front of the lens between its optical centre and principal focus.

Derive lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ .

A point object $O$ is placed in front of a glass rod having a spherical end of radius of curvature $30 c m$ . The image would be formed at :

A lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I$ . Aperture of diameter $\frac{d}{2}$ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively.

The distance between an object and the screen is $100 c m$ . A lens produces an image on the screen when the lens is placed at either of the positions $40 c m$ apart. The power of the lens is nearly

A ray incident at a point at an angle of incidence of $60^{o}$ enters a glass sphere of $R . l . n = 3$ and is reflected and refracted at the further surface of the sphere. The angel between the reflected and refracted rays at this surface is

Principle Rays For Lens

Language

Rate

REVISE WITH CONCEPTS

Convenient Rays to Draw Ray Diagrams

Image Formation by Lenses

QUICK SUMMARY WITH STORIES

Convenient Rays to Draw Ray Diagrams

Problem-based on Image Formation by Lens of Moving Object.

Derive the Lens Maker's formula.

A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such away that the end closer to the pole is 20 cm away from it. Find the length of the image.

One half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Explain your observations.

Derive Lens Maker's formula for a convex lens.

Draw a ray diagram to show the formation of image by a convex lens when an object is placed in front of the lens between its optical centre and principal focus.

Derive lens formula v1​−u1​=f1​ .

A point object O is placed in front of a glass rod having a spherical end of radius of curvature 30 cm. The image would be formed at :

A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter 2d​ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively.

The distance between an object and the screen is 100 cm. A lens produces an image on the screen when the lens is placed at either of the positions 40 cm apart. The power of the lens is nearly

A ray incident at a point at an angle of incidence of 60o enters a glass sphere of R.l.n=3​ and is reflected and refracted at the further surface of the sphere. The angel between the reflected and refracted rays at this surface is

Derive lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ .

A point object $O$ is placed in front of a glass rod having a spherical end of radius of curvature $30 c m$ . The image would be formed at :

A lens having focal length $f$ and aperture of diameter $d$ forms an image of intensity $I$ . Aperture of diameter $\frac{d}{2}$ in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively.

The distance between an object and the screen is $100 c m$ . A lens produces an image on the screen when the lens is placed at either of the positions $40 c m$ apart. The power of the lens is nearly

A ray incident at a point at an angle of incidence of $60^{o}$ enters a glass sphere of $R . l . n = 3$ and is reflected and refracted at the further surface of the sphere. The angel between the reflected and refracted rays at this surface is