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

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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Object and Lens Placed in Water

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}\$

Q: 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.

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)\$

Physics

example

Focal Length of a combination of lenses immersed in a medium

Example: The two surfaces of a biconvex lens has same radii of curvatures. This lens is made of glass of refractive index

1.5

and has a focal length

10 c m

in air. The lens is cut into two equal halves along a plane perpendicular to its principal axis to yield two plano-convex lenses. Then they are put in contact with their convex surfaces touching each other. If this combination lens is immersed in water (refractive index

= \frac{4}{3}

), then find its focal length (in

c m

).
Solution:

μ_{2} = 1.5

and

f_{a i r} = 10 c m

f

of 2 plano convex lenses

= f

of a biconvex lens

= 2 f_{a i r} = 20 c m

So,

\frac{f _{a i r}}{f _{l i q}}

\frac{\frac{μ _{g}}{μ _{l}} - 1}{μ _{g} - 1}

\frac{1 . 1 2 5 - 1}{1 . 5 - 1}

And,

f_{l i q} = 80 c m

f_{e q} = \frac{f _{1} f _{2}}{f _{1} + f _{2}} = \frac{8 0 \times 8 0}{1 0 0} = 40 c m

example

Example using lens formula

An object is situated at a distance of 15cm from a convex lens of focal length 30 cm. The position of the image formed by it will be

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\frac{1}{v} - \frac{1}{- 1 5} = \frac{1}{3 0}

\frac{1}{v} = \frac{1}{3 0} - \frac{1}{1 5}

\frac{1}{v} = \frac{- 1}{3 0}

v = - 30

REVISE WITH CONCEPTS

Calculation of Focal Length for Combination of Mirror and Lens

ExampleDefinitionsFormulaes

Problems in Lenses and Mirrors - Problem L1

ExampleDefinitionsFormulaes

Important Questions

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.

Medium

View solution

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?

Hard

View solution

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

Medium

View solution

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.

Medium

View solution

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?

Medium

View solution

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

Medium

View solution

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

Hard

View solution

A conversing lens has a focal length of $20 c m$ 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.

Hard

View solution

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

Medium

View solution

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.

Medium

View solution

example

Focal Length of a combination of lenses immersed in a medium

example

Example using lens formula

REVISE WITH CONCEPTS

Calculation of Focal Length for Combination of Mirror and Lens

Problems in Lenses and Mirrors - Problem L1

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 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 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.

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?

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

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

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.

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

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.

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 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

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

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 conversing lens has a focal length of $20 c m$ 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.

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