REFRACTION AT SPHERICAL SURFACES AND BY LENSES -

Refraction at Spherical Surfaces (introduction)

In the last two topics, so far we have only seen refraction and its application by plane surfaces. Now in this topic Refraction at Spherical Surfaces and by lenses, we will study at the refraction by a spherical surface. In this topic, we will also cover PHOTOMETRY, which is in your N.C.E.R.T syllabus. Finally, we will study about refraction by a lens and its applications. So here we are going to discuss the topic Refraction at Spherical Surfaces and by lenses. Following points I am going to cover which is mention below in the yellow box.

SPHERICAL REFRACTING SURFACE

A surface which forms a part of a sphere of the transparent medium is called a spherical refracting surface.

There are two types of spherical refracting surface.

1. CONVEX SPHERICAL SURFACE:-

A Spherical refracting surface which is convex towards the rarer medium is called convex spherical surface.

2. CONCAVE SPHERICAL SURFACE:-

A spherical refracting surface which is concave towards the rarer medium is called concave spherical surface.

REFRACTION AT CONVEX SPHERICAL SURFACE FROM RARER TO DENSER MEDIUM

Let XY is a convex spherical surface, separates rarer medium from a denser medium

Let P – pole of surface
C – center of curvature

Consider a point object O lying on the principal axis. After refraction through surface XY, a real image I is formed on a denser medium as shown in fig. from A, draw AM perpendicular on the principal axis.

Let $\fn_cm \angle AOM=\alpha , \angle AIM=\beta , and \angle ACM=\gamma$

from fig, $\fn_cm \tan \alpha =\frac{AM}{MO}$

if $\fn_cm \alpha$ is small (for paraxial ray)

$\fn_cm \large \tan \alpha \approx \alpha$

$\fn_cm \therefore \alpha =\frac{AM}{MO}-----(a)$

similarly,

$\fn_cm \beta =\frac{AM}{MI}-----(b)$

$\fn_cm \gamma =\frac{AM}{MC}-----(c)$

We know that for any, the external angle is the sum of two opposite internal angles.

i.e from $\fn_cm \Delta AOC$

$\fn_cm \large i=\alpha +\gamma -----(1)$

& from $\fn_cm \Delta AIC,$

$\fn_cm \large \gamma =r+\beta$

$\fn_cm \large r=\gamma -\beta -----(2)$

according to Snell’s law

$\fn_cm \large \frac{\sin i}{\cos r}=\frac{\mu _{2}}{\mu _{1}}$

for small angle, we can write

$\fn_cm \large \frac{i}{r}=\frac{\mu _{2}}{\mu _{1}}$

from (1) and (2)

$\fn_cm \large \frac{\alpha +\gamma }{\gamma -\beta }=\frac{\mu _{2}}{\mu _{1}}$

from (a), (b) and (c), we get

$\fn_cm \large \frac{\frac{AM}{MO}+\frac{AM}{MC}}{\frac{AM}{MC}-\frac{AM}{MI}}=\frac{\mu _{2}}{\mu _{1}}$

$\fn_cm \large \frac{\frac{1}{MO}+\frac{1}{MC}}{\frac{1}{MC}-\frac{1}{MI}}=\frac{\mu _{2}}{\mu _{1}}$

for paraxial ray, P and M are coincide

i.e $\fn_cm \large \frac{\frac{1}{PO}+\frac{1}{PC}}{\frac{1}{PC}-\frac{1}{PI}}=\frac{\mu _{2}}{\mu _{1}}$

using sign convention

$\fn_cm PO=-u,PC=R,PI=v$

$\fn_cm \large \therefore \frac{\frac{1}{-u}+\frac{1}{R}}{\frac{1}{R}-\frac{1}{v}}=\frac{\mu _{2}}{\mu _{1}}$

$\fn_cm \large \Rightarrow -\frac{\mu _{1}}{u}+\frac{\mu _{1}}{R}=\frac{\mu _{2}}{R}+\frac{\mu _{2}}{v}$

$\fn_cm \LARGE \Rightarrow \left [\frac{\mu _{2}}{v}-\frac{\mu _{1}}{u}=\frac{\mu _{2}-\mu _{1}}{R} \right ]$

REFRACTION AT CONCAVE SPHERICAL SURFACE FROM RARER TO DENSER MEDIUM

Students do it by themselves, just like I did above. I am just making a ray diagram here.

REFRACTION AT CONVEX SPHERICAL SURFACE FROM DENSER TO RARER MEDIUM

Do this in exactly the same way.

REFRACTION AT CONCAVE SPHERICAL SURFACE FROM DENSER TO RARER MEDIUM

Do this in exactly the same way.

PHOTOMETRY

It is known that a body above absolute zero temperature emits Electromagnetic radiation. The wavelength region in which the body emits the radiation depends on its absolute temperature.

For example, a Tungsten filament lamp having temperature 2850 K are partly invisible and mostly in IR (or heat) region.

The radiation capable of producing visible sensation to the human eye is called light energy or luminous energy. Only a small part ( $\fn_cm \small \lambda =4000 A^{0}$ to $\fn_cm \small 8000 A^{0}$ ) represents the visible or luminous energy

The measurement of light as perceived by the human eye is called photometry.

The branch of physics which deals with the measurement of light is called photometry.

LUMINOUS FLUX( $\fn_cm \large \phi$ ):-

The luminous flux from a light source is the luminous energy (i.e visible energy) emitted per second by the source. It is denoted by $\fn_cm \large \phi$ . Its arbitrary unit is lumen (lm)

for green light ( $\fn_cm \small f=540\times 10^{12}HZ$ )

$\fn_cm \small 1 lumen=\frac{1}{683}watt$

$\fn_cm \small =0.00146 watt$

NOTE- A standard 100 watts incandescent light bulb emits approx 1700 lumen

The dimension of luminous flux ( $\fn_cm \large \phi$ ) is $\fn_cm \left [ ML^{2}T^{-3} \right ]$

LUMINOUS INTENSITY OR ILLUMINATING POWER (I):-

the luminous intensity of a light source is the light radiating capacity of the source in a given direction. It is defined as ” The luminous intensity of a light source in any direction is the luminous flux emitted by the source per unit solid angle in that direction.” It is denoted by I.

ie. $\fn_cm \large \left [ I=\frac{\phi }{\omega } \right ]$

where $\fn_cm \large \phi \rightarrow$ luminous flux of $\fn_cm \large \phi$ lumens.

$\fn_cm \large \omega \rightarrow$ solid angle in steradians

Its S.I unit is lumens/ steradian (lm/sr) which is also called candela (cd) or candle power (C.P). Its dimension is $\fn_cm \large \left [ ML^{2}T^{-3} \right ]$

Now, consider a point light source of luminous intensity I. If we draw a sphere around the point source of light, the sphere would subtend a solid angle of $\fn_cm \large 4\Pi$ steradian at the source. If is the luminous flux emitted by the light source, then

$\fn_cm \large \left [ I=\frac{\phi }{4\Pi } \right ]$

$\fn_cm \large \left [ \phi =4\Pi I \right ]$

ILLUMINANCE OF SURFACE (E):-

When luminous flux falls on a surface, the surface is said to be illuminated. Therefore illuminance is always for a surface where the light falls. It is defined as luminous flux falling normally on the unit area of the surface. It is denoted by E.

i.e $\fn_cm \large \left [ E=\frac{\phi }{A} \right ]$

Its S.I unit is $\fn_cm lumen/m^{^{2}}$ . (lux/meter candela)

Now we know that $\fn_cm E=\frac{\phi }{A}$

for point source $\fn_cm E=\frac{4\Pi I}{4\Pi r^{2}}$

$\fn_cm \left [ E=\frac{I}{r^{2}} \right ]$

If I is constant, then

$\fn_cm \left [ E\propto \frac{1}{r^{2}} \right ]$

i.e the illuminance (E) of a surface is inversely proportional to the square of the distance (r) between the light source and the surface.

LUMINANCE (OR BRIGHTNESS):-

The luminance of a surface refers to the brightness of the surface and is given by the luminous flux reflected from a unit area of the surface. It depends on:

a. Illuminance of the surface i.e the luminous flux incident on a unit area of the surface and

b. nature of the surface

Its unit is $\fn_cm cd/m^{2}$ (sometimes called ‘nit’ in the industry)

A good LCD monitor has a brightness of about 250 nits.

LENSES

A lens is a portion of a transparent refracting medium bounded by two spherical surfaces or one spherical and the other plane surface.

we deal only thin lens.

There are two types of lens

1. Convex or Converging lens:- It converges parallel rays.

2. Concave or Diverging lens:-It diverges parallel rays.

IMPORTANT TERM RELATED TO LENSES

LENS MAKER’S FORMULA

It is so-called because it enables the manufacturers of lenses to design the lenses of required focal length from a glass of given R.I.

It is the relation between the focal length of the lens, the radii of curvature of its two surfaces the R.I of the material of lens and the R.I of the surrounding.

Consider a thin convex lens XY of optical center P. Let $R_{1}$ and $R_{2}$ is the radius of curvature of two surfaces. let $\mu _{1}$ be the R.I of the lens and $\mu _{2}$ is the R.I of the surrounding.

Consider a point object O is placed on the principal axis. After 1st refraction a virtual image I’ is formed.

for 1st refraction (Curved surface)

$\fn_cm \large \frac{\mu _{2}}{v'}-\frac{\mu _{1}}{u}=\frac{\mu _{2}-\mu _{1}}{R_{1}}-----(1)$

(NOTE:- sign convention is not used here, because at the time of this derivation we already took sign convention)

for 2nd refraction, Image I’ serves as a purpose of an object, and after 2nd refraction a real image I is formed on the principal axis.

i.e $\fn_cm \large \frac{\mu _{1}}{v}-\frac{\mu _{2}}{v'}=\frac{\mu _{1}-\mu _{2}}{R_{2}}-----(2)$

adding (1) and (2) we get

$\fn_cm \large \frac{\mu _{1}}{v}-\frac{\mu _{1}}{u}=\left ( \mu _{2}-\mu _{1} \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )$

$\fn_cm \large \frac{1}{v}-\frac{1}{u}=\left ( \frac{\mu _{2}}{\mu _{1}}-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )$

$\fn_cm \large if\; u\rightarrow \infty \; ,\; then\; v\rightarrow f$

$\fn_cm \large \left [ \frac{1}{f}=\left ( \frac{\mu _{2}}{\mu _{1}}-1 \right )\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right ) \right ]$

This is the lens maker’s formula.

NOTE:-

1. If rays coming parallel to the principal axis of the lens, then after refraction it passes through the principal focus or appears to diverge from it.

2. A ray through the optical centre passes undeviated because the middle of the lens acts like a thin parallel slab.

LENS FORMULA

It is the relation between object distance, image distance and focal length.

consider xy is a convex lens of focal length f. AB is an object placed perpendicular to the principal axis. After refraction, a real, inverted image A’B’ is formed on the principal axis.

from fig. $\Delta BPA\; \; and \; \; \Delta B'PA'\; \; \; are\; \; \; similar$

$\fn_cm i.e \; \; \frac{B'A'}{BA}=\frac{PA'}{PA}-----(1)$

$also \; \; \Delta MFP\; \; \; and\; \; \; \Delta B'FA'\; \; \; are \; \; similar$

$\fn_cm i.e\; \; \frac{B'A'}{MP}=\frac{FA'}{PF}$

$\fn_cm \because MP=BA$

$\fn_cm \therefore \frac{B'A'}{BA}=\frac{FA'}{PF}-----(2)$

$\therefore \; \; from (1)\; \; and \; \; (2)$

$\fn_cm \Rightarrow \frac{PA'}{PA}=\frac{FA'}{PF}$

$\fn_cm \Rightarrow \frac{PA'}{PA}=\frac{PA'-PF}{PF}$

using sign convention

$\fn_cm PA=-u,\; \; PA'=v\; \; and \; \; PF=f$

$\fn_cm \large \therefore \frac{v}{-u}=\frac{v-f}{f}$

$\fn_cm \large \Rightarrow vf=-uv+uf$

divide both sides by uvf, we get

$\fn_cm \large \Rightarrow \frac{vf}{uvf}=\frac{-uv}{uvf}+\frac{uf}{uvf}$

$\fn_cm \large \Rightarrow \frac{1}{u}=-\frac{1}{f}+\frac{1}{v}$

$\fn_cm \large \left [ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \right ]$ which is lens formula

LINEAR MAGNIFICATION

It is the ratio of image height to the object height, It is denoted by m.

$\fn_cm \large i.e\; \;\; \left [ m=\frac{h_{2}}{h_{1}} \right ]$

where $\large h_{2}$ – height of the image

$\large h_{1}$ – height of the object

$\fn_cm from\; above\;figure \;\; \frac{B'A'}{BA}=\frac{PA'}{PA}$

$\fn_cm \large \Rightarrow \frac{-h_{2}}{h_{1}}=\frac{v}{-u}$

$\fn_cm \large \left [ m=\frac{v}{u} \right ]$

IMAGE FORMATION BY LENS

1. FOR CONVEX LENS

2. FOR CONCAVE LENS

POWER OF LENS

The power of a lens is the ability of the lens to bend the path of rays passing through it.

i.e The greater the power of a lens, the greater is its ability to bend the light rays falling on it & vice-versa.

Hence smaller the focal length of a lens more is its ability to bend light rays and greater is its power.

i.e The power of the lens is measured as the reciprocal of its focal length. It is denoted by P.

$\fn_cm \large i.e \; \; \left [ P=\frac{1}{f} \right ]$

its S.I unit is /m or dioptre (D)

The power of a convex lens is +ve and that of a concave lens is -ve.

NOTE:-

It is also said that ” The power of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light falling at unit distance from the optical centre.

$\fn_cm \large tan\theta =\frac{h}{f}$

$\fn_cm \large for \;small\;angle\;tan\theta \approx \theta$

$\fn_cm \large i.e\;\theta =\frac{h}{f}$

$\fn_cm \large if\;h=1,\; then\;\theta =\frac{1}{f}$

$\fn_cm \large thus\; \left [ p=\frac{1}{f} \right ]$

COMBINATION OF THIN LENSES ( IN CONTACT)

Consider two thin convex lenses A and B of focal length $\fn_cm \large f_{1}\;and \;f_{2}$ placed co-axially in contact with each other.

Suppose a point object O is placed on principal axis at a distance u from the lenses.

for lens A,

O is an object and I’ is the real image at a distance v’ from A.

i.e we can write

$\fn_cm \large \frac{1}{v'}-\frac{1}{u}=\frac{1}{f_{1}}-----(1)$

This image I’ serves as a virtual object for lens B.

i.e for lens B, I’ is a virtual object and I is the real image at a distance v from B.

i.e we can write

$\fn_cm \large \frac{1}{v}-\frac{1}{v'}=\frac{1}{f_{2}}-----(2)$

adding (1) and (2) we get,

$\fn_cm \large \frac{1}{v}-\frac{1}{u}=\frac{1}{f_{1}}+\frac{1}{f_{2}}-----(a)$

If we replace the two lenses by a single lens of focal length f which forms the image I at a distance v for object O at distance u from it, then this lens is known as an equivalent lens.

i.e we can write for equivalent lens

$\fn_cm \large \frac{1}{v}-\frac{1}{u}=\frac{1}{f}-----(b)$

from (a) and (b)

$\fn_cm \large \left [ \frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}} \right ]$

equivalent power is

$\fn_cm \large \left [ p=p_{1}+p_{2} \right ]$

where $\fn_cm \large p_{1}\;and\;p_{2}$ are the power of lens A and B respectively.

NOTE-

1. The net magnification, when two lenses of linear magnification $\fn_cm \large m_1\;and\;m_{2}$ are in contact is

$\fn_cm \large \left [ m_{net}=m_{1}m_{2} \right ]$

2. We know that

$\fn_cm \large \frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}$

If one lens is convex and other is concave, then we can write

$\fn_cm \large \frac{1}{f}=\frac{1}{f_{1}}-\frac{1}{f_{2}}$

$\fn_cm \large \frac{1}{f}=\frac{f_{2}-f_{1}}{f_{1}f_{2}}$

$\fn_cm \large \therefore \left [ f=\frac{f_{1}f_{2}}{f_{2}-f_{1}} \right ]$

case I – If $\fn_cm \large f_{1}=f_{2}$ , then $\fn_cm \large f\rightarrow \infty$ , this means that the combination would behave like a plane glass plate.

case II- If $\fn_cm \large f_{1}>f_{2}$ , then $\fn_cm \large f$ is -ve, i.e the combination acts as a concave lens

case III- If $\fn_cm \large f_{1}<f_{2}$ , then $\fn_cm \large f$ is +ve, i.e the combination acts as a convex lens

3. If two lenses of focal length $\fn_cm \large f_{1}\;and\;f_{2}$ are separated by a finite distance d, then it can be proved that the focal length f of the equivalent lens is given by.

$\fn_cm \large \left [ \frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{d}{f_{1}f_{2}} \right ]$

4. A lens made of three different materials have three focal lengths. i.e for a given object there are three images.

7. Power of lens in the liquid is given by $\fn_cm \large \left [ p=\frac{\mu _{liquid}}{f} \right ]$

8. The focal length of the mirror depends on the only radius $\fn_cm \large f=\frac{R}{2}$ . But for the lens, it depends upon $\fn_cm \large \mu_{1}, \mu_{2}, R{1}\;and \; R_{2}$

We know that

$\fn_cm \large \frac{1}{f}=(\frac{\mu_{2} }{\mu_{1} }-1)\left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )$

if $\fn_cm \large \mu_{1}> \mu_{2}$ , then f is -ve. The behaviour of the lens is changed. i.e converging lens behaves as a diverging lens and vice-versa.

ex- bubble inside water.

10. During a refraction, a ray of light doesn’t bend under the following two conditions:

a. If the light is incident normally (i=0)

b. if $\fn_cm \large \mu_{1}=\mu_{2}$

RELATED LINK OF RAY OPTICS AND OPTICAL INSTRUMENTS
Introduction	Refraction
Total Internal Reflection	Reflection of light by Spherical Mirror
Refraction through a Prism	Dispersion by a Prism
Some Natural Phenomena due to Sunlight	Optical Instruments

IMPORTANT QUESTION FOR PRACTICE

1. Light from a point source in the air falls on a spherical glass surface of R.I=1.5 and radius of curvature=20 cm. The distance of the light source from the glass surface is 100 cm. At what position the image is formed?

2. A magician during a show makes a glass lens with R.I= 1.47 disappear in a trough of liquid. What is the R.I of the liquid? Could the liquid be water?

3. (i) if f=0.5 m for a glass lens, what is the power of the lens? (ii) The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive index of glass? (iii) A convex has 20 cm focal length in air. What is the focal length in the air? What is the focal length in the water? ( Refractive index of air-water=1.33, refractive index for air-glass=1.5.)

4. Find the position of the image formed by the lens combination given in the following figure.

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

6. 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?

7. An object of size 3 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?

8. What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore the thickness of the lenses.

9. The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

10. A screen is placed 90 cm 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. Determine the focal length of the lens.

11. (a) Determine the ‘effective focal length’ of the combination of the two lenses in Ex- 8, if they are placed 8 cm apart with their principal axis coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of the effective focal length of this system useful at all?

(b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system and the size of the image.

12. Following fig shows a biconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30 cm. What is the refractive index of the liquid?

Refraction at Spherical Surfaces (introduction)

SPHERICAL REFRACTING SURFACE

There are two types of spherical refracting surface.

1. CONVEX SPHERICAL SURFACE:-

2. CONCAVE SPHERICAL SURFACE:-

REFRACTION AT CONVEX SPHERICAL SURFACE FROM RARER TO DENSER MEDIUM

REFRACTION AT CONCAVE SPHERICAL SURFACE FROM RARER TO DENSER MEDIUM

REFRACTION AT CONVEX SPHERICAL SURFACE FROM DENSER TO RARER MEDIUM

REFRACTION AT CONCAVE SPHERICAL SURFACE FROM DENSER TO RARER MEDIUM

PHOTOMETRY

LUMINOUS FLUX():-

LUMINOUS INTENSITY OR ILLUMINATING POWER (I):-

ILLUMINANCE OF SURFACE (E):-

LUMINANCE (OR BRIGHTNESS):-

LENSES

There are two types of lens

IMPORTANT TERM RELATED TO LENSES

LENS MAKER’S FORMULA

NOTE:-

LENS FORMULA

LINEAR MAGNIFICATION

IMAGE FORMATION BY LENS

1. FOR CONVEX LENS

2. FOR CONCAVE LENS

POWER OF LENS

NOTE:-

COMBINATION OF THIN LENSES ( IN CONTACT)

NOTE-

IMPORTANT QUESTION FOR PRACTICE

LUMINOUS FLUX( $\fn_cm \large \phi$ ):-