Q: A concave lens of focal length 15cm forms an image 10 cm from the lens . How far is the object placed from the lens? Draw the ray diagram.

Given: \$f=-15\$ \$v=-10\$From mirror formula, \$\cfrac{1}{v}-\cfrac{1}{u}=\cfrac{1}{f}\$ \$\cfrac{1}{-10}-\cfrac{1}{u}=\cfrac{1}{-15}\$ \$\cfrac{1}{u}=\cfrac{1}{15}-\cfrac{1}{10}\$ \$u=-30\ cm\$

Q: A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam \$12\$ cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length \$20\$ cm, and (b) a concave lens of focal length \$16\$ cm?

\$(a)\$. According to the given case, object is virtual, \$u=+12\ cm\$The focal length of the convex lens is \$f=+20\ cm\$.Using lens equation, \$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\$\$\dfrac{1}{v}-\dfrac{1}{12}=\dfrac{1}{20}\$\$v=7.5\ cm\$\$(b)\$. Now the focal length of the concave lens, \$f=-16\ cm\$\$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\$\$\dfrac{1}{v}-\dfrac{1}{12}=\dfrac{1}{-16}\$\$v=48\ cm\$

Q: A convex lens \$A\$ of focal length \$20cm\$ and a concave lens \$B\$ of focal length \$5cm\$ are kept along the same axis with the distance \$d\$ between them. If a prallel beam of light falling on \$A\$ leaves \$B\$ as a parallel beam, then the distance \$d\$ in cm will be

\$d={f}_{1}-{f}_{2}\$By using the formula\$\cfrac { 1 }{ { f }_{ 1 } } \quad +\cfrac { 1 }{ { f }_{ 1 } } -\cfrac { d }{ { f }_{ 1 }{ f }_{ 2 } } =\cfrac { 1 }{ F } \$For the emergent beam to parallel\$P=0,F=\alpha \$\$\Rightarrow \cfrac { 1 }{ 20 } -\cfrac { 1 }{ 5 } +\cfrac { d }{ 100 } =0\$\$\Rightarrow\$ \$d=15\$

Q: An object is kept in front of a concave mirror of focal length 20 cm. The image is three times the size of the object.Calculate two possible distance of the object from the mirror.

It is given that, Focal length \$ f=-20\,cm \$ \$ {{h}_{i}}=3{{h}_{o}} \$ The image is three times the size of the object. So there are two cases. Case 1. For real image \$ m=-\dfrac{v}{u}=-3 \$ \$ v=3u \$ Using mirror’s formula \$ \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} \$ \$ \dfrac{1}{-20}=\dfrac{1}{3u}+\dfrac{1}{u} \$ \$ u=\dfrac{-80}{3}\,cm \$ Case 2. For virtual image \$ m=\dfrac{-v}{u}=+3 \$ \$ v=-3u \$ Again using mirror’s formula \$ \dfrac{1}{-20}=\dfrac{1}{-3u}+\dfrac{1}{u} \$ \$ u=\,\dfrac{-40}{3}\,cm \$ The possible values of u are \$\dfrac{-80}{3}\,cm\,\,and\,\,\dfrac{-40}{3}\,cm\$.

Q: Double convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?

Lens maker's formula,\$\dfrac{1}{f} \, = \, (\mu \, - \, 1) \left ( \dfrac{1}{R_1} \, - \, \dfrac{1}{R_2} \right )\$Here, f = 20 cm, \$\mu\$ = 1.55, \$R_1\$ = R, \$R_2\$ = -R\$\dfrac{1}{20} \, = \, (1.55 \, - \, 1) \left ( \dfrac{1}{R} \, - \, \dfrac{1}{(-R)} \right ) \,\, or \,\, \dfrac{1}{20} \, = \, 0.55 \, \times \, \dfrac{2}{R}\$\$\Rightarrow \, R \, = \, 1.1 \, \times \, 20 \, = \, 22 \, cm\$

Q: A ray of light falls on a transparent sphere with centre at \$C\$ is shown in Fig.The ray emerges from the sphere parallel to line \$AB\$. The refractive index of the sphere is

Deviation by a sphere is \$2(i - )\$Here, deviation \$\delta = 60^{\circ} = 2(i - r)\$or \$i - r = 30^{\circ}\$\$\therefore r = i - 30^{\circ} = 60^{\circ} - 30^{\circ} = 30^{\circ}\$\$\therefore \mu = \dfrac {\sin i}{\sin r} = \dfrac {\sin 60^{\circ}}{\sin 30^{\circ}} = \sqrt {3}\$.

Q: A conversing lens has a focal length of \$20\ cm\$ in air. It is made of a material of refractive index \$1.6\$. If it is immersed in a liquid of refractive index \$1.3\$, find the new focal length.

Converging lens of focal length = \$+20cm\$ in air \$\mu_L=1.6\$\$\dfrac{1}{F}=\left(\dfrac{\mu_L}{\mu_M}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)\$ = Lens formulaWhere \$\mu_L\rightarrow\$ refractive index of lens \$\mu_M\rightarrow\$ refractive index of mediumIn air \$(\mu_M=1)\$\$\dfrac{1}{20}=\left(\dfrac{1.6}{1}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)\$\$\dfrac{1}{20}=0.6\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)\$\$\dfrac{1}{12}=\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)\$In liquid \$(\mu_M=1.3)\$\$\dfrac{1}{F}=\left(\dfrac{1.6}{1.3}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)\$\$\dfrac{1}{F}=\dfrac{0.3}{1.3}\times \dfrac{1}{12}\$\$\dfrac{1}{F}=\dfrac{1}{52}\$Hence,new focal length is \$52\$ cm

Q: The focal length of a convex lens of glass \$(\mu =1.5)\$ is \$2cm\$. The focal length of the lens when immersed in a liquid of refractive index \$1.25\$ will be :

Casr \$1\$ lens in airLet the focal length of lens in air be fairGiven that: \$F_{air}\$ \$=2\ cm,\ n_1=1(air),\ n_2=1.5 (g\ case)\$\$\dfrac {1}{F_{air}}=\left [\dfrac {n_2}{n_1}-1\right] \left [\dfrac {1}{R_1}-\dfrac {1}{R_2}\right]\$\$\dfrac {1}{2}= \left [\dfrac {1.5}{1}-1\right]\left [\dfrac {1}{R_1} -\dfrac {1}{R_2}\right]\$\$\dfrac {1}{2}=0.5 \left [\dfrac {1}{R_1}-\dfrac {1}{R_2}\right].......(1)\$Case \$2\$ lens in water\$F_{water}\ \Rightarrow \ n_1=1.25,\ n_2=1.5\$\$\dfrac {1}{F_{water}}=\left [\dfrac {n_2}{n_1}-1\right]\left [\dfrac {1}{R_1}-\dfrac {1}{R_2}\right]\$\$\dfrac {1}{F_{water}}=\left [\dfrac {1.5}{1.25}-1\right]\left [\dfrac {1}{R_1}-\dfrac {1}{R_2}\right]\$\$\dfrac {1}{F_{water}}=0.2 \left [\dfrac {1}{R_1}-\dfrac {1}{R_2}\right].........(2)\$Dividing \$(1)\$ by \$(2)\$ we get\$\dfrac {F_{water}}{2}=\dfrac {0.5}{0.2}\$\$\boxed {\Rightarrow \ F_{water}=5\ cm}\$

Question 1

A concave lens of focal length 15cm forms an image 10 cm from the lens . How far is the object placed from the lens? Draw the ray diagram.

Accepted Answer

Given: $f=-15$            $v=-10$From mirror formula, $\cfrac{1}{v}-\cfrac{1}{u}=\cfrac{1}{f}$                                    $\cfrac{1}{-10}-\cfrac{1}{u}=\cfrac{1}{-15}$                                    $\cfrac{1}{u}=\cfrac{1}{15}-\cfrac{1}{10}$                                    $u=-30\ cm$

Question 2

A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam $12$ cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length $20$ cm, and (b) a concave lens of focal length $16$ cm?

Accepted Answer

$(a)$. According to the given case, object is virtual, $u=+12\ cm$The focal length of the convex lens is $f=+20\ cm$.Using lens equation, $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$$\dfrac{1}{v}-\dfrac{1}{12}=\dfrac{1}{20}$$v=7.5\ cm$$(b)$. Now the focal length of the concave lens, $f=-16\ cm$$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$$\dfrac{1}{v}-\dfrac{1}{12}=\dfrac{1}{-16}$$v=48\ cm$

Question 3

A convex lens $A$ of focal length $20cm$ and a concave lens $B$ of focal length $5cm$ are kept along the same axis with the distance $d$ between them. If a prallel beam of light falling on $A$ leaves $B$ as a parallel beam, then the distance $d$ in cm will be

Accepted Answer

$d={f}_{1}-{f}_{2}$By using the formula$\cfrac { 1 }{ { f }_{ 1 } } \quad +\cfrac { 1 }{ { f }_{ 1 } } -\cfrac { d }{ { f }_{ 1 }{ f }_{ 2 } } =\cfrac { 1 }{ F } $For the emergent beam to parallel$P=0,F=\alpha $$\Rightarrow \cfrac { 1 }{ 20 } -\cfrac { 1 }{ 5 } +\cfrac { d }{ 100 } =0$$\Rightarrow$ $d=15$

Question 4

An object is kept in front of a concave mirror of focal length 20 cm. The image is three times the size of the object.Calculate two possible distance of the object from the mirror.

Accepted Answer

It is given that,&#10;&#10;Focal length&#10;&#10;$ f=-20\,cm $&#10;&#10;$ {{h}_{i}}=3{{h}_{o}} $&#10;&#10;The image is three times the size of the object.&#10;So there are two cases.&#10;&#10;Case 1. For real image&#10;&#10;$ m=-\dfrac{v}{u}=-3 $&#10;&#10;$ v=3u $&#10;&#10;Using mirror&#8217;s formula&#10;&#10;$ \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} $&#10;&#10;$ \dfrac{1}{-20}=\dfrac{1}{3u}+\dfrac{1}{u} $&#10;&#10;$ u=\dfrac{-80}{3}\,cm $&#10;&#10;Case 2. For virtual image&#10;&#10;$ m=\dfrac{-v}{u}=+3 $&#10;&#10;$ v=-3u $&#10;&#10;Again using mirror&#8217;s formula&#10;&#10;$ \dfrac{1}{-20}=\dfrac{1}{-3u}+\dfrac{1}{u} $&#10;&#10;$ u=\,\dfrac{-40}{3}\,cm $&#10;&#10;The possible values of u are $\dfrac{-80}{3}\,cm\,\,and\,\,\dfrac{-40}{3}\,cm$.

Question 5

Double convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?

Accepted Answer

Lens maker's formula,$\dfrac{1}{f} \, = \, (\mu \, - \, 1) \left ( \dfrac{1}{R_1} \, - \, \dfrac{1}{R_2} \right )$Here, f = 20 cm, $\mu$ = 1.55, $R_1$ = R, $R_2$ = -R$\dfrac{1}{20} \, = \, (1.55 \, - \, 1) \left ( \dfrac{1}{R} \, - \, \dfrac{1}{(-R)} \right ) \,\, or \,\, \dfrac{1}{20} \, = \, 0.55 \, \times \, \dfrac{2}{R}$$\Rightarrow \, R \, = \, 1.1 \, \times \, 20 \, = \, 22 \, cm$

Question 6

Derive the following expression for the refraction at concave spherical surface: $\displaystyle\frac{\mu}{v}-\frac{1}{u}=\frac{\mu -1}{R}$.

Accepted Answer

Let MPN be concabe spherical surface of a medium of refractive index $\mu$. P is the pole o is the centre of curvature PC is the principle axis. Let O be a point object or it's principle axis kept in a rarer medium first incident ray travelling through C is normal to the spherical surface therefor it will not deviate and goes along PX another mident ray OA will refract at Point A and it bends towards the normal. AB and PX are proceeded behind then at I, a virtual image will be form.Let $\alpha, \beta, \gamma$ are the ray angle normal with the principle axis respectively.then by snell's law$\mu =\displaystyle\frac{sin i}{sin r}$            ........(1)but here i and r are the small then we put.$sin i=i$       $sin r=r$ in equation (1)$\mu =\displaystyle\frac{i}{r}$$i=\mu n$                 ...........(2)now by using exterior angle theorem in $\Delta$AOC$\gamma =1+a$$i=\gamma -\alpha$             .........(3)now in $\Delta$IAC by using exterior angle theorem$\gamma =b+r$$\rho =\gamma -\beta$            ..........(4)putting value of i and r in equation (2)$(\gamma -\alpha)=\mu(\gamma -\beta)$           ...........(5)now angle$=\displaystyle\frac{arc}{radius}$$\alpha =\displaystyle\frac{PA}{OP}$$\beta =\frac{PA}{IP}$now $\gamma =\displaystyle\frac{PA}{CP}$using value of $\alpha, \beta, \gamma$ in equation (5)$\displaystyle\frac{PA}{PC}-\frac{PA}{PO}=\mu\left(\displaystyle\frac{PA}{PC}-\frac{PA}{PI}\right)$$PA\left(\displaystyle\frac{1}{PC}-\frac{1}{PO}\right)=m. PA\left(\displaystyle\frac{1}{PC}-\frac{1}{PO}\right)$$\displaystyle\frac{1}{PC}-\frac{1}{PO}=\mu\left(\displaystyle\frac{1}{PC}-\frac{1}{PI}\right)$         .........(6)now by using sign convention$PC=-R$                $PI=V$$PO=-u$using these value in equation (6)$\left(\displaystyle\frac{1}{-R}\right)-\left(\displaystyle\frac{1}{-u}\right)=\mu \left(\displaystyle\frac{-1}{R}+\frac{1}{V}\right)$$\displaystyle\frac{-1}{R}+\frac{1}{u}=\mu\left(\displaystyle\frac{-1}{R}+\frac{1}{v}\right)$$\displaystyle\frac{-1}{R}+\frac{1}{u}=\frac{\mu}{R}+\frac{\mu}{v}$$\displaystyle\frac{-1}{R}+\frac{\mu}{R}=\frac{\mu}{v}-\frac{1}{u}$$\displaystyle\frac{\mu -1}{R}=\frac{\mu}{v}-\frac{1}{u}$This is the required expression for the refraction formula for the concave spherical surface.

Question 7

A ray of light falls on a transparent sphere with centre at $C$ is shown in Fig.The ray emerges from the sphere parallel to line $AB$. The refractive index of the sphere is

Accepted Answer

Deviation by a sphere is $2(i - )$Here, deviation $\delta = 60^{\circ} = 2(i - r)$or $i - r = 30^{\circ}$$\therefore r = i - 30^{\circ} = 60^{\circ} - 30^{\circ} = 30^{\circ}$$\therefore \mu = \dfrac {\sin i}{\sin r} = \dfrac {\sin 60^{\circ}}{\sin 30^{\circ}} = \sqrt {3}$.

Question 8

A conversing lens has a focal length of $20\ cm$ in air. It is made of a material of refractive index $1.6$. If it is immersed in a liquid of refractive index $1.3$, find the new focal length.

Accepted Answer

Converging lens of focal length = $+20cm$ in air $\mu_L=1.6$$\dfrac{1}{F}=\left(\dfrac{\mu_L}{\mu_M}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$ = Lens formulaWhere $\mu_L\rightarrow$ refractive index of lens            $\mu_M\rightarrow$ refractive index of mediumIn air $(\mu_M=1)$$\dfrac{1}{20}=\left(\dfrac{1.6}{1}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$$\dfrac{1}{20}=0.6\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$$\dfrac{1}{12}=\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$In liquid $(\mu_M=1.3)$$\dfrac{1}{F}=\left(\dfrac{1.6}{1.3}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$$\dfrac{1}{F}=\dfrac{0.3}{1.3}\times \dfrac{1}{12}$$\dfrac{1}{F}=\dfrac{1}{52}$Hence,new focal length is $52$ cm

Question 9

The focal length of a convex lens of glass $(\mu =1.5)$ is $2cm$. The focal length of the lens when immersed in a liquid of refractive index $1.25$ will be :

Accepted Answer

Casr $1$ lens in airLet the focal length of lens in air be fairGiven that: $F_{air}$ $=2\ cm,\ n_1=1(air),\ n_2=1.5 (g\ case)$$\dfrac {1}{F_{air}}=\left [\dfrac {n_2}{n_1}-1\right] \left [\dfrac {1}{R_1}-\dfrac {1}{R_2}\right]$$\dfrac {1}{2}= \left [\dfrac {1.5}{1}-1\right]\left [\dfrac {1}{R_1} -\dfrac {1}{R_2}\right]$$\dfrac {1}{2}=0.5 \left [\dfrac {1}{R_1}-\dfrac {1}{R_2}\right].......(1)$Case $2$ lens in water$F_{water}\ \Rightarrow \ n_1=1.25,\ n_2=1.5$$\dfrac {1}{F_{water}}=\left [\dfrac {n_2}{n_1}-1\right]\left [\dfrac {1}{R_1}-\dfrac {1}{R_2}\right]$$\dfrac {1}{F_{water}}=\left [\dfrac {1.5}{1.25}-1\right]\left [\dfrac {1}{R_1}-\dfrac {1}{R_2}\right]$$\dfrac {1}{F_{water}}=0.2 \left [\dfrac {1}{R_1}-\dfrac {1}{R_2}\right].........(2)$Dividing $(1)$ by $(2)$ we get$\dfrac {F_{water}}{2}=\dfrac {0.5}{0.2}$$\boxed {\Rightarrow \ F_{water}=5\ cm}$

Question 10

A screen is placed 90 cm away from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Find the focal length of lens.

Accepted Answer

The image of the object can be located on the screen for two-position of convex lens such that u and v are exchanged.The separation between two positions of the lens is, d= 20 cmFrom the figure,$u_1 + v_1 = 90 cm$$v_1 - u_1 =20 cm$Solving,$v_1 = 55 cm, \  u_1 = 35 cm$From lens formula,$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$ $\dfrac{1}{55} - \dfrac{1}{-35} = \dfrac{1}{f}$$\Rightarrow \dfrac{1}{f}=\dfrac{1}{55} + \dfrac{1}{35} = \dfrac {55\times 35}{90} $$\Rightarrow f= 21.4cm$Hence, the correct option is $(B)$

Combination of Lenses and Mirrors

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Calculation of Focal Length for Combination of Mirror and Lens

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