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 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: 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: 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: The minimum distance between an object and its real image formed by a convex lens is :

Let the between object and real image be \$d\$. Let the distance of object from the lens is \$x\$So the distance of image and lens be \$d-x\$ Using the lens equation :\$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\$\$\dfrac{1}{f}=\dfrac{1}{d-x}-\dfrac{1}{-d}\$\$x=\dfrac{d\pm \sqrt{d^2-4fd}}{2}\$ ...(1)So for real \$x\$\$\sqrt{d^2-4fd}\ge0\$So \$d\ge 4f\$So minimum distance between object and image is 4f.

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: An object is placed 50 cm from a lens produces a virtual image at a distance of 10 cm in front of the lens. Draw a diagram to show the formation of image and calculate the focal length of the lens.

Distance of object from the mirror \$u=-50\$ cmDistance of image from the mirror \$v=-10\$ cmFor lens, we know\$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\$\$\dfrac{1}{f}=\dfrac{1}{-10}-\dfrac{1}{-50}\$\$f=-12.5cm\$It is a concave lens.

Q: A biconvex lens has a radius of curvature of magnitude \$20\ cm\$. Which one of the following options best describe the image formed of an object of height \$2\ cm\$ placed \$30\ cm\$ from the lens?

Given, Radius of curvature, \$R = 20\ cm\$Height of object, \$h_{0} = 2\$Object distance, \$u = 30\ cm\$We have, \$\dfrac {1}{f} = (\mu - 1) \left (\dfrac {1}{R_{1}} - \dfrac {1}{R_{2}}\right )\$\$= \left (\dfrac {3}{2} - 1\right ) \left [\dfrac {1}{20} - \left (-\dfrac {1}{20}\right )\right ]\$\$\Rightarrow \dfrac {1}{f} = \left (\dfrac {3}{2} - 1\right ) \times \dfrac {2}{20}\$\$\therefore f = 20\ cm\$\$\dfrac {1}{f} = \dfrac {1}{v} - \dfrac {1}{u}\$\$\Rightarrow \dfrac {1}{20} = \dfrac {1}{v} + \dfrac {1}{30}\$\$\dfrac {1}{v} = \dfrac {1}{20} -\dfrac {1}{30}\$\$= \dfrac {10}{600}\$\$v = 60\ cm\$\$m = \dfrac {h_{i}}{h_{0}} = \dfrac {v}{u}\$\$\Rightarrow h_{i} = \dfrac {v}{u} \times h_{0}\$\$= \dfrac {60}{30}\times 2 = -4\ cm\$so, image is inverted.

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 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 3

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 4

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 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

The minimum distance between an object and its real image formed by a convex lens is :

Accepted Answer

Let the between object and real image be $d$. Let the distance of object from the lens is $x$So the distance of image and lens be $d-x$  Using the lens equation :$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$$\dfrac{1}{f}=\dfrac{1}{d-x}-\dfrac{1}{-d}$$x=\dfrac{d\pm  \sqrt{d^2-4fd}}{2}$  ...(1)So for real $x$$\sqrt{d^2-4fd}\ge0$So $d\ge 4f$So minimum distance between object and image is 4f.

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

An object is placed 50 cm from a lens produces a virtual image at a distance of 10 cm in front of the lens. Draw a diagram to show the formation of image and calculate the focal length of the lens.

Accepted Answer

Distance of object from the mirror $u=-50$ cmDistance of image from the mirror $v=-10$ cmFor lens, we know$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$$\dfrac{1}{f}=\dfrac{1}{-10}-\dfrac{1}{-50}$$f=-12.5cm$It is a concave lens.

Question 9

A biconvex lens has a radius of curvature of magnitude $20\ cm$. Which one of the following options best describe the image formed of an object of height $2\ cm$ placed $30\ cm$ from the lens?

Accepted Answer

Given, Radius of curvature, $R = 20\ cm$Height of object, $h_{0} = 2$Object distance, $u = 30\ cm$We have, $\dfrac {1}{f} = (\mu - 1) \left (\dfrac {1}{R_{1}} - \dfrac {1}{R_{2}}\right )$$= \left (\dfrac {3}{2} - 1\right ) \left [\dfrac {1}{20} - \left (-\dfrac {1}{20}\right )\right ]$$\Rightarrow \dfrac {1}{f} = \left (\dfrac {3}{2} - 1\right ) \times \dfrac {2}{20}$$\therefore f = 20\ cm$$\dfrac {1}{f} = \dfrac {1}{v} - \dfrac {1}{u}$$\Rightarrow \dfrac {1}{20} = \dfrac {1}{v} + \dfrac {1}{30}$$\dfrac {1}{v} = \dfrac {1}{20} -\dfrac {1}{30}$$= \dfrac {10}{600}$$v = 60\ cm$$m = \dfrac {h_{i}}{h_{0}} = \dfrac {v}{u}$$\Rightarrow h_{i} = \dfrac {v}{u} \times h_{0}$$= \dfrac {60}{30}\times 2 = -4\ cm$so, image is inverted.

Question 10

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.

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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.

A convex lens $A$ of focal length $20 c m$ and a concave lens $B$ of focal length $5 c m$ 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

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.

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?

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?

The minimum distance between an object and its real image formed by a convex lens is :

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 $A B$ . The refractive index of the sphere is

An object is placed 50 cm from a lens produces a virtual image at a distance of 10 cm in front of the lens. Draw a diagram to show the formation of image and calculate the focal length of the lens.

A biconvex lens has a radius of curvature of magnitude $20 c m$ . Which one of the following options best describe the image formed of an object of height $2 c m$ placed $30 c m$ from the lens?

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

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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.

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

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.

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?

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?

The minimum distance between an object and its real image formed by a convex lens is :

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

An object is placed 50 cm from a lens produces a virtual image at a distance of 10 cm in front of the lens. Draw a diagram to show the formation of image and calculate the focal length of the lens.

A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options best describe the image formed of an object of height 2 cm placed 30 cm from the lens?

Derive the following expression for the refraction at concave spherical surface: vμ​−u1​=Rμ−1​.

A convex lens $A$ of focal length $20 c m$ and a concave lens $B$ of focal length $5 c m$ 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

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 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 $A B$ . The refractive index of the sphere is

A biconvex lens has a radius of curvature of magnitude $20 c m$ . Which one of the following options best describe the image formed of an object of height $2 c m$ placed $30 c m$ from the lens?

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