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\$

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\$

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\$

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\$.

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\$

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.

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}\$.

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

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}\$

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)\$

Problems in Lenses and Mirrors - Problem L1 | Formulas, Definition, Examples

example

Image distance or object distance for refraction of rays at spherical surfaces

Where would an object be placed in a medium of refractive index

μ_{1}

, so that its real image is formed at equidistant from sphere of radius R and refractive index

μ_{2}

which is also placed in the medium of refractive index

μ_{1}

as shown in figure?
We know,

\frac{μ _{1}}{v} - \frac{μ _{2}}{u} = \frac{μ _{2} - μ _{1}}{R}

For 1, we get:

\frac{μ _{2}}{v} - \frac{μ _{1}}{- d} = \frac{( μ _{2} - μ _{1} )}{R}

- d = u =

object distance
or,

\frac{μ _{1}}{v} = \frac{μ _{2} - μ _{1}}{R} - \frac{μ _{1}}{d} = t

For 2, we get:

\frac{μ _{1}}{d} - \frac{μ _{2}}{( 2 R - v )} = \frac{μ _{2} - μ _{1}}{R}

or,

(\frac{μ _{2} - μ _{1}}{R} - \frac{μ _{1}}{d}) = \frac{μ _{2}}{( 2 R - v )} = \frac{μ _{2}}{v}

or,

2 R - v = v

or,

R = v

So,

\frac{μ _{1}}{d} = \frac{μ _{2} - μ _{1}}{r}

or,

d = (\frac{μ _{1}}{μ _{2} - μ _{1}}) R

formula

Refraction of light at spherical surfaces

Let,

$n_{1}$ - refractive index of medium from which rays incident.

$n_{2}$ - refractive index of another medium.

u - distance of object from pole of spherical surface

v - distance of image from pole of spherical surface

$t a n α = \frac{M N}{O M}$

$t a n γ = \frac{M N}{M C}$

$t a n β = \frac{M N}{M I}$

Now, for $Δ$ NOC, i is the exterior angle.

$i = ∠ N O M + ∠ N C M$

$i = \frac{M N}{O M} + \frac{M N}{M C}$ .........(1)

Similarly,

$r = \frac{M N}{M C} - \frac{M N}{M I}$ .........(2)

Now by using snells law we get

$n_{1} s i n i = n_{2} s i n r$

Or for small angles

$n_{1} i = n_{2} r$

Substituting i and r from eq. (1) and (2), we get

$\frac{n _{1}}{O M} + \frac{n _{2}}{M I} = \frac{n _{2} - n _{1}}{M C}$

As,

OM = -u, MI = +v, MC = + R

Hence equation becomes

$\frac{n _{2}}{v} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}$

example

Minimum distance between object and its real image for a convex lens

Example: In a convex lens of focal length

f

, find the minimum distance between an object and its real image.

Solution:
Let image distance be

v

, object distance be

- u

and focal length be

f

.
Let distance between object and image be

s

∴ s = u + v

Now,

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

or,

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

or,

\frac{s}{u ( s - u )} = \frac{1}{f}

or,

s = \frac{u ^{2}}{u - f}

For point of minima,

\frac{d s}{d u} = 0

or,

\frac{u ^{2} - 2 u f}{( u - f ) ^{2}} = 0

or,

u = 0, 2 f

Therefore,

s = \frac{( 2 f ) ^{2}}{2 f - f} = 4 f

definition

Magnification

Magnification of a lens is defined as:

m = \frac{I}{O} = \frac{v}{u}

Note:
Sign convention must be followed while using formula for magnification. Hence, it can be positive or negative.

∣ m ∣ > 1 ⟹

image is magnified.

∣ m ∣ = 1 ⟹

image is same size as object.

∣ m ∣ < 1 ⟹

image is diminished.

example

Calculate Focal Length of Lenses using lens maker's formula

Example: A plano-convex lens has thickness

4 c m

. When placed on a horizontal table with the curved surface in contact with it, the apparent depth of the bottom-most point of the lens is found to be

3 c m

. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of the plane face of the lens is found to be

\frac{2 5}{8} c m

. Find the focal length of the lens.

Solution:
When the curved surface of the lens (refractive index

μ

) is in contact with the table, the image of the bottom-most point of lens (in glass) is formed due to refraction at plane face.
The image of O appears at

I_{1}

.
Here,

u_{1} = A O = - 4 c m, v_{1} = A I_{1} = 3 c m

μ_{1} = μ

μ_{2} = 1

, and

R_{1} = \infty

∴ \frac{μ _{2}}{v _{1}} - \frac{μ _{1}}{u _{1}} = \frac{μ _{2} - μ _{1}}{R _{1}}

gives,

\frac{1}{- 3} - \frac{μ}{- 4} = \frac{1 - μ}{\infty}

................ (i)
When the plane surface of the lens in contact with the table, the image of center of the plane face is formed due to refraction at curved surface. The image of O is formed at

I_{2}

.
Here,

u_{2} = A O = - 4 c m, v_{2} = A I_{2} = - \frac{2 5}{8} c m

μ_{1} = μ, μ_{2} = 1

, and

R_{2} = - R

∴ \frac{μ _{2}}{v _{2}} - \frac{μ _{1}}{u _{2}} = \frac{μ _{2} - μ _{1}}{R _{2}}

gives.

\frac{1}{( - \frac{2 5}{8} )} - \frac{μ}{- 4} = \frac{1 - μ}{- R}

From Eq. (i), we get:

μ = 4 / 3

, therefore this equation gives

- \frac{8}{2 5} + \frac{4 / 3}{4} = - \frac{( 1 - \frac{4}{3} )}{R} = - \frac{8}{2 5} + \frac{1}{3} = \frac{1}{3 R}

\frac{1}{7 5} = \frac{1}{3 R}

This gives

R = 25 c m

The focal length (

f

) of plano-convex lens

(R_{1} = R

and

R_{2} = \infty)

\frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{\infty}) = \frac{μ - 1}{R} = \frac{\frac{4}{3} - 1}{2 5} = \frac{1}{7 5}

\Rightarrow f = 75 c m

example

Calculate magnification of lens

Problem:
The focal length of a thin biconvex lens is

20 c m

. When an object is moved from a distance of

25 c m

in front of it to

50 c m

, the magnification of its image changes from

m_{25}

m_{50}

. The ratio

\frac{m _{25}}{m _{50}}

is :
Solution:
When object is at

25 c m

u = - 25 c m

v = 100 c m

m_{25} = \frac{v}{u} = \frac{1 0 0}{2 5}

.....(1)

When object is at

50 c m

u = - 50 c m

v = 100 / 3 c m

m_{50} = \frac{v}{u} = \frac{1 0 0}{1 5 0}

.....(2)

We have to find

\frac{m _{25}}{m _{50}}

\frac{m _{25}}{m _{50}} = \frac{\frac{1 0 0}{2 5}}{\frac{1 0 0}{1 5 0}} = 6

example

The minimum distance between an object and its real image in a convex lens

Distance between object and image

= 2 f + 2 f = 4 f

Problems in Lenses and Mirrors - Problem L1

example

Image distance or object distance for refraction of rays at spherical surfaces

formula

Refraction of light at spherical surfaces

example

Minimum distance between object and its real image for a convex lens

definition

Magnification

example

Calculate Focal Length of Lenses using lens maker's formula

example

Calculate magnification of lens

example

The minimum distance between an object and its real image in a convex lens

REVISE WITH CONCEPTS

Calculation of Focal Length for Combination of Mirror and Lens

Object and Lens Placed in Water

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 $20 c m$ 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.

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: $\frac{μ}{v} - \frac{1}{u} = \frac{μ - 1}{R}$ .

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 :

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.

example

Image distance or object distance for refraction of rays at spherical surfaces

formula

Refraction of light at spherical surfaces

example

Minimum distance between object and its real image for a convex lens

definition

Magnification

example

Calculate Focal Length of Lenses using lens maker's formula

example

Calculate magnification of lens

example

The minimum distance between an object and its real image in a convex lens

REVISE WITH CONCEPTS

Calculation of Focal Length for Combination of Mirror and Lens

Object and Lens Placed in Water

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 :