Optics Concept Page - 10

Result

Experimental determination of refractive index of glass

Using the apparatus shown in the diagram, direct a ray of light to enter the block near the middle of the longest side and to leave by the opposite side crossing directly from one side to the other. As you change the angle of the block to the light, notice the alterations in the direction of the emerging ray.

• Draw round the edges of the block for one arrangement.
• Mark the paths of the light outside the block with a few pencil dots.
• Remove the block and draw in (with a ruler) the path of the ray through the block as a straight line.
• Use arrows to show which way the light travelled.
• Construct a normal where the ray enters the block and measure i and r.

Repeat for at least five different angles of incidence. Calculate the refractive index for the material of the glass block using Snell's law.
Refractive index, n=sinisinr$n=\frac{\sin i}{\sin r}$

Law

Relation between refractive index and speed of light

Refractive index of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in air.
μ or n=cv$\mu \text{ or }n={\displaystyle \frac{c}{v}}$

The speed at which light propagates through materials such as glass or air, is less than c; just as the speed of radio waves in wire cables is slower than c.
The refractive index is the relationship between the speed of light and the speed at which light travels. The refractive index of light in glass is about 1.5 which means that light is 1.5 slower in that medium.
The light changes direction when is incident to another medium at an angle, otherwise, you will not notice.

Example

Refractive index of some common mediums

The Refractive Index n$n$ of a medium is a measure for how much the velocity of light is reduced inside the medium. The velocity of light in a medium can be expressed as

v=cn$v={\displaystyle \frac{c}{n}}$

Material	Refractive index w.r.t vacuum
Vacuum	1
Air	1.0003
Water	1.33
Ordinary glass	1.5
Diamond	2.41

Example

Refractive index in a medium with varying refractive index

Example: The index of refraction of a glass plate is 1.48$1.48$ at  θ1=30oC${\theta }_1={30}^{o}C$ and varies linearly with temperature with a coefficient of 2.5×10−5$2.5\times {10}^{-5}$ oC−1${{}^{o}C}^{-1}$.  The coefficient of linear expansion of the glass is 5×10−5$5\times {10}^{-5}$ oC−1${{}^{o}C}^{-1}$. At 30oC${30}^{o}C$, the length of the glass plate is 3cm$3cm$. This plate is placed in front of one of the slits in Young's double-slit temperature increases at a rate of 5oC−1$5^o{C}^{-1}$min, the light source has wavelength λ=589 nm$\lambda =589nm$ and the glass plate initially is at θ=30oC$\theta ={30}^{o}C$. Find the number of fringes that shift on the screen in each minute. (use approximation)

Solution:
Path Difference due to slab at t=0$t=0$ is given by:
δslab=t×(1+αΔT)×(μ×(1+βΔT)−1)${\delta }_{slab}=t\times (1+\alpha \mathrm{\Delta }T)\times (\mu \times (1+\beta \mathrm{\Delta }T)-1)$
Here, α=2.5×10−5/0C$\alpha =2.5\times {10}^{-5}{/}^{0}C$, β=5×10−5/0C$\beta =5\times {10}^{-5}{/}^{0}C$ and ΔT=5oCmin−1$\mathrm{\Delta }T=5^{o}Cmi{n}^{-1}$
Distance between each fringe = nλDd$n\lambda \frac{D}{d}$
Therefore equating both values for a complete minute:
Δδslab=t×(μ×(1+αΔT+βΔT+αβΔT2)−αΔT−1)=nλ$\mathrm{\Delta }{\delta }_{slab}=t\times (\mu \times (1+\alpha \mathrm{\Delta }T+\beta \mathrm{\Delta }T+\alpha \beta \mathrm{\Delta }{T}^2)-\alpha \mathrm{\Delta }T-1)=n\lambda$, where Δδslab=δt=1min−δt=0$\mathrm{\Delta }{\delta }_{slab}={\delta }_{t=1min}-{\delta }_{t=0}$
⇒$\Rightarrow$ n≈11$n\approx 11$

Definition

Terms related to refraction

Incident ray: The ray which falls on the surface of separation(or interface) to enter into the new medium.
Refracted ray: The ray in the second medium, obtained after refraction. 
Normal: Imaginary straight line perpendicular to the refracting surface at the point of refraction.
Angle of incidence (i)$(i)$: Angle between the incident ray and the normal.
Angle of refraction (r)$(r)$: Angle between the refracted ray and the normal.

Law

Laws of refraction

According to laws of refraction (Snell's Laws):

1.  The incident ray, the refracted ray and the normal at the point of incidence, all lie in the same plane.
2. The ratio of the sine of the angle of incidence i$i$ to the sine of the angle of refraction is constant for the pair of given media. This constant is called the refractive index of the second medium w.r.t. the first medium.
   1μ2=sin isin r${}_1{\mu }_2={\displaystyle \frac{sini}{sinr}}$

Note:

• When light ray is incident normally, only speed changes and direction of  light remains the same.
• When light ray passes from rarer medium to denser medium, it bends towards the normal.
• When light ray passes from denser medium to rarer medium, it bends away from the normal.

Example

Calculate refractive index of medium

Problem: 
A parallel beam of natural light is incident at an angle of 58∘${}^{\circ }$ on a plane glass surface. The reflected beam is completely linearly polarized(tan 58∘=${}^{\circ }=$1.6). The angle of refraction of the transmitted beam and the refractive index of the glass are :
Solution:

If the light in incident on the surface with an
angle of incidence i given by tan i =μ$=\mu$,
the reflected light is completely polarized.
Thus μ=tan580$\mu =tan{58}^0$
refractive μ=1.6$\mu =1.6$
index
From snell's law
sinisinγ=μ${\displaystyle \frac{sini}{sin\gamma }}=\mu$
sin5801.6=sinγ${\displaystyle \frac{sin{58}^0}{1.6}}=sin\gamma$
γ=320$\gamma ={32}^0$

Definition

Snell's Law

The ratio of the sine of the angle of incidence to the sine of angle of refraction is constant. This is known as the Snell's Law. 

From Snell's Law, we have

n1sinθ1=n2sinθ2$n_{1}sin{\theta }_1=n_{2}sin{\theta }_2$

hence,

sinθ1sinθ2=n2n1=v1v2=λ1λ2${\displaystyle \frac{sin{\theta }_1}{sin{\theta }_2}}={\displaystyle \frac{n_2}{n1}}={\displaystyle \frac{v_1}{v_2}}={\displaystyle \frac{{\lambda }_1}{{\lambda }_2}}$

Law

Experimental Verification of Law of refraction

The incident ray, the refracted ray and the normal to the surface at the point of incidence all lie in one plane. For any two given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant.

sinisinr=μ${\displaystyle \frac{sini}{sinr}}=\mu$

Where μ$\mu$is the refractive index of the second medium with respect to the first medium.

• Place a rectangular glass slab on the white sheet of paper fixed on a drawing board.
• Trace the boundary ABCD of the glass slab.
• Remove the glass slab and draw a normal N1N2 at O.
• Draw a straight line IO inclined at an angle say 300${30}^0$ with the normal. IO is the incident ray.
• Fix two pins P and Q on the incident ray IO.
• Place the glass slab within its boundary ABCD.
• Looking from the other side of the glass slab fix two other pins R and S such that P, Q, R and S appear to lie on the same straight line.

• Remove the glass slab and the pins. Mark the pin points P, Q, R and S.
• Join the pins R and S and produce the line on both sides. The ray O'E is the emergent ray.
• Join OO'. It is the refracted ray.
• The incident ray, the refracted ray and the normal are all lying in the same plane.
• This proves the first law of refraction.
• Let us now prove the second law of refraction
• With O as center, draw a circle of a convenient radius 'R' in such a way that it cuts the incident and the refracted rays at F and G respectively.
• From F and G draw perpendiculars to the normal N1N2.
• Triangle FHO and triangle GKO are right-angled triangles.

sini=FHOF$sini={\displaystyle \frac{FH}{OF}}$
sinr=GKOG$sinr={\displaystyle \frac{GK}{OG}}$

μ=sinisinr=FHOF×OGGK$\mu ={\displaystyle \frac{sini}{sinr}}={\displaystyle \frac{FH}{OF}}\times {\displaystyle \frac{OG}{GK}}$
But OG=OF=R$OG=OF=R$

μ=FHOG×OGGK$\mu ={\displaystyle \frac{FH}{OG}}\times {\displaystyle \frac{OG}{GK}}$
μ=FHGK$\mu ={\displaystyle \frac{FH}{GK}}$

• Measure the length of FH and GK and record them in the observation table.
• Repeat the experiment for different values of angle of incidence.
• Find the value of FHGK${\displaystyle \frac{FH}{GK}}$ for different values of i.
• In each case it is found that the ratio FHGK${\displaystyle \frac{FH}{GK}}$ is the same, that is, the ratio of FHGK${\displaystyle \frac{FH}{GK}}$ is a constant. This verifies law of refraction.

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