REFRACTION -

Refraction of light (introduction)

Refraction of Light can be seen in many places in our everyday life. There is a convex lens in our eyes, where the light from the object, being refracted by the lens of the eye form an image on the retina. It makes objects under a water surface appears closer than they really are. Refraction is also responsible for some natural phenomena such as rainbows, sparkling of diamond, mirage etc. Refraction is used in the working of the telescope, microscope, Cameras, Magnifying glass etc. So here we are going to discuss the topic Refraction of Light. Following points I am going to cover which is mention below in the yellow box.

REFRACTION OF LIGHT

The refraction of light is the phenomenon of change in the path of light when it passes from one medium to another medium.

In going from a rarer medium to denser medium, the ray bends towards the normal, and in going from denser to rarer medium, the ray bends away from the normal.

Q> What is the basic cause of refraction?

Ans> It is due to a change in velocity of light in going from one medium to the another.

(You can understand it, very well in wave optics.)

REFRACTIVE INDEX

It gives the optical density of the medium.

Higher optical density means higher Refractive Index
Lower optical density means lower Refractive Index

The refractive index of a medium is a measure of how much the speed of light is reduced when it enters that medium.

i.e we can say that it is the property of medium due to which, when the ray passes through it, its velocity is changed.

It is the ratio of the speed of light in a vacuum to the speed of light in a medium.

it is denoted by $\fn_cm \LARGE \mu$

i.e $\fn_cm \large \left [ \mu =\frac{speed\: of\: light\: in \: vacuume}{speed\: of \: light\: in\displaystyle \: medium} \right ]$

NOTE:-

1. Optical density should not be confused with a mass density which is mass per unit volume, it is possible that the mass density of an optically denser medium may be less than that of an optically rarer medium.

for example, turpentine and water. The mass density of turpentine is less than that of water but its optical density is higher.

2. when a ray passes from one medium to another, its frequency, the phase is not changed, but its velocity, path, and wavelength is changed.

3. $\fn_cm \large \left [ v=\nu \lambda \right ] or \left [ v=f\lambda \right ] or \left [ v=n\lambda \right ]$

where $\fn_cm \large v$ – the speed of light, $\fn_cm \large f/\nu /n$ – frequency of light and $\fn_cm \large \lambda$ – wavelength of light

4. $\fn_cm \large \left [ \mu \propto \frac{1}{v} \right ]$

5. R.I of vacuum/air=1

R.I of water= 4/3=1.33
R.I of glass= 3/2=1.5

6. R.I of the medium is given always w.r.t another medium

for example $\fn_cm \large \mu _{21}$ or $\fn_cm \large _{1}\mu _{2}$ = R.I of second medium w.r.t first medium

7. If nothing is given, the R.I of the medium is always w.r.t air (vacuum), which is known as absolute R.I

e.x- $\fn_cm \large \mu _{water}=1.33$ this is w.r.t air/ vacuum

8. $\fn_cm \large \left [ \mu _{21}=\frac{\mu _{2}}{\mu _{1}} \right ]$

LAWS OF REFRACTION OF LIGHT

There are two laws of refraction of light

1. The incident ray, the refracted ray, and the normal at the point of incidence, all lie in one plane.

2. The ratio of the sine of the angle of incidence to the sine of the angle of the refraction is constant

i.e $\fn_cm \large \left [ \frac{\sin i}{\sin r}=\frac{\mu _{2}}{\mu _{1}}=constant \right ]$ This is known as Snell’s law.

REFRACTION THROUGH DIFFERENT MEDIUM ( FOR PARALLEL SLAB)

using Snell’s law for all three interfaces and multiply all, we get

$\fn_cm \large \frac{ \sin i}{ \sin r_{1}}\times \frac{ \sin r_{1}}{ \sin r_{1}}\times \frac{ \sin r_{2}}{ \sin r}=\frac{\mu _{2}}{\mu _{1}}\times \frac{\mu _{3}}{\mu _{2}}\times \frac{\mu _{1}}{\mu _{3}}$

$\fn_cm \large \frac{\sin i}{ \sin r}=1$

$\fn_cm \LARGE \left [ i=r \right ]$

i.e emergent ray and incident ray are parallel if both outer medium is the same.

REAL AND APPARENT DEPTH

An object placed in a denser medium, when viewed from a rarer medium, appears to be at a lesser depth than its real depth. This is an account due to the refraction of light.

From fig
AO- Real depth
AI- Apparent depth which is less than real depth

now from $\fn_cm \Delta AOB$

$\fn_cm \tan i=\frac{AB}{AO}$

for paraxial approximation $\fn_cm \tan i \approx \sin i$

$\fn_cm \therefore \sin i=\frac{AB}{AO}$

similarly, from $\fn_cm \Delta AIB$

$\fn_cm \tan r=\frac{AB}{AI}$

for paraxial approximation $\fn_cm \tan r\approx \sin r$

$\fn_cm \therefore \sin r=\frac{AB}{AI}$

from Snell’s law

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

$\fn_cm \frac{AB}{AO}\times \frac{AI}{AB}=\frac{\mu _{1}}{\mu _{2}}$

$\fn_cm \frac{AI}{AO}=\frac{\mu _{1}}{\mu _{2}}$

$\fn_cm AI=\left ( \frac{\mu _{1}}{\mu _{2}} \right )AO$

If the rarer medium is air ( $\fn_cm \mu_{1} =1$ ) and the other medium is ( $\fn_cm \mu _{2}=\mu$ ), then we can write

$\fn_cm \large \left [ AI=\frac{AO}{\mu } \right ]$ i.e $\fn_cm \large \left [ Apparent\: Depth=\frac{Real\; Depth}{\mu } \right ]$

REAL AND APPARENT HEIGHT

An object placed in a rarer medium, when viewed from a denser medium, appears to be at a greater height than its real height. This is an account due to the refraction of light.

From fig
AO- Real height
AI- Apparent height which is more than real height

now from $\fn_cm \Delta AOB$

$\fn_cm \tan i=\frac{AB}{AO}$

for paraxial approximation $\fn_cm \tan i \approx \sin i$

$\fn_cm \therefore \sin i=\frac{AB}{AO}$

similarly, from $\fn_cm \Delta AIB$

$\fn_cm \tan r=\frac{AB}{AI}$

for paraxial approximation $\fn_cm \tan r\approx \sin r$

$\fn_cm \therefore \sin r=\frac{AB}{AI}$

from Snell’s law

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

$\fn_cm \frac{AB}{AO}\times \frac{AI}{AB}=\frac{\mu _{2}}{\mu _{1}}$

$\fn_cm \frac{AI}{AO}=\frac{\mu _{2}}{\mu _{1}}$

$\fn_cm AI=\left ( \frac{\mu _{2}}{\mu _{1}} \right )AO$

If the rarer medium is air ( $\fn_cm \mu_{1} =1$ ) and the other medium is ( $\fn_cm \mu _{2}=\mu$ ), then we can write

$\fn_cm \large \left [ AI=\mu \cdot AO \right ]$

i.e $\fn_cm \large \left [ Apparent\; Height=\mu \cdot Real\; Height \right ]$

SHIFT DUE TO GLASS SLAB

NORMAL SHIFT

Consider an object O is placed in front of the glass slab of the refractive index $\fn_cm \large \mu$ . For the first refraction, O is an object and I’ is its virtual image. This image I’ acts as an object for the second refraction which finally forms the virtual image at I as shown in fig, so shifting occurs from O to I. From fig OI is called the normal shift and its value is.

$\fn_cm \left [ OI=(1-\frac{1}{\mu })t \right ]$

Let $\fn_cm \large AO=x$

$\fn_cm \large \therefore AI'=\mu x$

$\fn_cm \large \Rightarrow BI'=\mu x+t$

$\fn_cm \large \Rightarrow BI=\frac{\mu x+t}{\mu }=x+\frac{t}{\mu }$

$\fn_cm \large \therefore OI=(AB+AO)-BI$

$\fn_cm \large =(t+x)-(x+\frac{t}{\mu })$

$\fn_cm \large =t-\frac{t}{\mu }$

$\fn_cm \large \left [ OI=(1-\frac{1}{\mu })t \right ]$

NOTE:- The direction of the shift is actually in the direction of the incident light.

LATERAL SHIFT

When a ray of light passes through a rectangular glass slab, the emergent ray is parallel to the incident ray, although there is lateral displacement. The perpendicular distance between the incident and emergent rays is called lateral displacement.

From fig, $\fn_cm \large d$ is the lateral shift and $\fn_cm \large t$ is the thickness of the glass slab.

let $\fn_cm \angle BOD=(i-r)=$ Deviation on refraction at the first surface

in $\fn_cm \Delta BOD$

$\fn_cm \sin (i-r)=\frac{BD}{OB}$

$\fn_cm \therefore BD=OB\sin (i-r)-------------(1)$

in $\fn_cm \Delta OEB$

$\fn_cm \cos r=\frac{OE}{OB}$

$\fn_cm \therefore OB=\frac{OE}{\cos r}=\frac{t}{ \cos r}$

$\fn_cm \therefore$ from (1) $\fn_cm BD=\frac{t}{\cos r}\cdot \sin (i-r)$

$\fn_cm \therefore \left [ Lateral\; Displacement(d)=BD=\frac{t\sin (i-r)}{\cos r} \right ]$

NOTE:-

1. Lateral displacement increases with

a. thickness of glass slab

b. The angle of incidence

c. Refractive index of the slab

2. Lateral displacement will be maximum when $\fn_cm i=90^{0}$

ATMOSPHERIC REFRACTION

REFRACTION EFFECTS AT SUNRISE AND SUNSET

Due to atmospheric refraction, the sun is visible before the actual sunrise and sunset.

from fig, S is the actual position of the sun which is just below the horizon during sunrise/sunset. When rays from the sun enter the earth’s atmosphere, they meet a medium of increasing density, as a result, they bend slightly towards the normal, and the sun appears at position S’ (apparent position), which is just above the horizon by the observer on earth.

The R.I if air w.r.t vacuum is 1.00029. Due to this, the apparent shift in the direction of the sun is by about $\fn_cm \large \frac{1}{2}^{0}$ and the corresponding time difference between actual sunset and apparent sunset is about 2 minutes. As the same way, the sun appears to rise early by about 2 minutes before the actual rise.

Therefore, the days become longer by about 4 minutes due to atmospheric refraction. ” The apparent flattening (oval shape) of the sun at sunset and sunrise is also due to the same phenomenon”

TWINKLING OF STAR

When the rays of light from the star enter the earth’s atmosphere, they meet a medium of increasing density. As a result, the rays of light continuously bend. Since the layers of air are not stationary due to convection current, the image of the star keeps changing its position and star twinkling.

Q> Why planet are not twinkling?

Ans> It is because the amount of light from them is quite large so the variation is not noticeable.

LIGHT TRAVEL IN SHORTEST TIME TAKEN PATH(LIFEGUARD AND SNELL’S LAW)

Consider a rectangular swimming pole PQRS. From fig. A lifeguard sitting at G outside the pool notice a child drowning at a point C. The guard wants to reach the child in the shortest possible time. Let SR be the side of the pool between G and C. Should he/she take a straight line path GAC or GBC or some other path GXC.

The guard knows that his/her running speed $\fn_cm \large v_{1}$ on ground and $\fn_cm \large v_{2}$ swimming speed on the water.

Suppose the guard enters the water at X.

Let, $\fn_cm \large GX=l_{1}$ , $\fn_cm \large XC=l_{2}$ , $\fn_cm \large BD=h$ , $\fn_cm \large BC=y$ , $\fn_cm \large GD=a$ , $\fn_cm \large GM=x$ , $\fn_cm \large MD= (a-x)$

$\fn_cm \large \angle GXM=i$ and $\fn_cm \large \angle CXL=r$

The time taken to reach from G to C would be

$\fn_cm \large t=\frac{l_{1}}{v_{_{1}}}+\frac{l_{2}}{v_{_{2}}}$

$\fn_cm \large t=\frac{\sqrt{x^{2}+h^{2}}}{v_{1}}+\frac{\sqrt{(a-x)^{2}+y^{2}}}{v_{2}}$

To make this time minimum, one has to differentiate it w.r.t x and it has to be equal to zero.

$\fn_cm \large i.e \; \: \: \frac{\mathrm{d} t}{\mathrm{d} x}=0$

$\fn_cm \large \frac{1}{v_{1}}\cdot \frac{\mathrm{d} \sqrt{x^{2}+h^{2}}}{\mathrm{d} \left ( x^{2}+h^{2} \right )}\cdot \frac{\mathrm{d} \left ( x^{2}+h^{2} \right )}{\mathrm{d} x}+\frac{1}{v_{2}}\cdot \frac{\mathrm{d} \sqrt{(a-x)^{2}+y^{2}}}{\mathrm{d} \left ( (a-x)^{2}+y^{2} \right )}\cdot \frac{\mathrm{d} \left ( (a-x)^{2}+y^{2} \right )}{\mathrm{d} x}=0$

$\fn_cm \large \Rightarrow \frac{1}{v_{1}}\cdot \frac{1}{2\sqrt{x^{2}+h^{2}}}\cdot 2x+\frac{1}{v_{2}}\cdot \frac{1}{2\sqrt{(a-x)^{2}+y^{2}}}\cdot 2\left \{ -(a-x) \right \}=0$

$\fn_cm \large \Rightarrow \frac{1}{v_{1}}\cdot \frac{x}{\sqrt{x^{2}+h^{2}}}=\frac{1}{v_{2}}\cdot \frac{(a-x)}{\sqrt{(a-x)^{2}+y^{2}}}$

from fig,

$\fn_cm \large \Rightarrow \frac{\sin i}{v_{1}}=\frac{\sin r}{v_{2}}$

$\fn_cm \large \left [ \frac{\sin i}{ \sin r} =\frac{v_{1}}{v_{2}} \right ]$

If $\fn_cm \large \mu _{2}$ is the R.I of water and $\fn_cm \large \mu _{1}$ is the R.I of air, and we know that $\fn_cm \large v\propto \frac{1}{\mu }$

i.e we can write

$\fn_cm \large \left [ \frac{\sin i}{ \sin r} =\frac{\mu _{2}}{\mu _{1}} \right ]$

which is just Snell’s law

i.e Whether it is a wave or a particle or a human being, whenever two medium and two velocities are involved, one must follow Snell’s law, if we want to take the shortest time.

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

IMPORTANT QUESTION FOR PRACTICE

1. A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?

2. The earth takes 24 hours to rotate once about its axis. How much time does the sun take to shift by $\fn_cm 1^{0}$ when viewed from the earth?

3. Fig show refraction of a ray in air incident at $\fn_cm 60^{0}$ with the normal to a glass- air and water-air interface, respectively. Predict the angle of refraction in a glass when the angle of incidence in water is $\fn_cm 45^{0}$ with the normal to a water – glass interface.

4. A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive Index of water is 1.33. (Consider bulb to be a point source)

5. A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? R.I of glass =1.5. Does the answer depend on the location of the slab?

6. Answer the following question:

a. A driver under water looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?

b. Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?