Optics Concept Page - 12

Example

Apparent depth of an object through layers of multiple media

Example: A vessel of depth d$d$ is filled with a liquid of refractive index μ1${\mu }_1$ up to half its depth and the remaining space is filled with a liquid of refractive index μ2${\mu }_2$. Find the apparent depth while seeing normal to the free surface of the liquid.

Solution:
The shift in depth due to first liquid =d2(1−1μ1)$={\displaystyle \frac{d}{2}}(1-{\displaystyle \frac{1}{{\mu }_1}})$
The shift in depth due to second liquid =d2(1−1μ2)$={\displaystyle \frac{d}{2}}(1-{\displaystyle \frac{1}{{\mu }_2}})$
So, apparent depth =d−d+d2(1μ1+1μ2)$=d-d+{\displaystyle \frac{d}{2}}({\displaystyle \frac{1}{{\mu }_1}}+{\displaystyle \frac{1}{{\mu }_2}})$ =d2(1μ1+1μ2)$={\displaystyle \frac{d}{2}}({\displaystyle \frac{1}{{\mu }_1}}+{\displaystyle \frac{1}{{\mu }_2}})$

Example

Calculate lateral displacement of glass slab having some thickness

Problem:
Refractive index of a rectangular glass slab is m=3√$m=\sqrt{3}$ . A light ray incident at an angle 60∘${60}^{\circ }$ is displaced laterally through 2.5cm$2.5cm$. Distance traveled by light in the slab is :
Solution:
Given μ=3√$\mu =\sqrt{3}$
θa=600${\theta }_a={60}^0$Lateral shift d=tsin(θa−θ′b)cosθ′b$d=t{\displaystyle \frac{\sin ({\theta }_a-{\theta }_b^{\prime })}{\cos {\theta }_b^{\prime }}}$ ...................(1)
Let the length traveled in glass be l$l$.
From geometry, t=lcosθ′b$t=l\cos {\theta }_b^{\prime }$ ..........................(2)
Substituting t$t$  from (2) in (1) d=lsin(θa−θ′b)$d=lsin({\theta }_a-{\theta }_b^{\prime })$ .............................................(3)
From Snell's law, sinθasinθ′b=μ${\displaystyle \frac{\sin {\theta }_a}{\sin {\theta }_b^{\prime }}}=\mu$
⇒sin600=3√sinθ′b$\Rightarrow \sin {60}^0=\sqrt{3}\sin {\theta }_b^{\prime }$
⇒sinθ′b=12$\Rightarrow \sin {\theta }_b^{\prime }={\displaystyle \frac{1}{2}}$
⇒θ′b=300$\Rightarrow {\theta }_b^{\prime }={30}^0$
Hence, from (3) 2.5=lsin(600−300)$2.5=l\sin ({60}^0-{30}^0)$
⇒l=5cm$\Rightarrow l=5cm$

Formula

Lateral displacement of light from glass slab

A ray of light strikes a glass slab of thickness t. It emerges on the opposite face, parallel to the incident ray but laterally displaced. The lateral displacement is  found asthe angle of deviation is i−r$i-r$
lateral displacement is BC(x)$BC(x)$(from figure)
x=BCsin(i−r)$x=BCsin(i-r)$
BC=ADcosr$BC={\displaystyle \frac{AD}{cosr}}$
AD=t$AD=t$(thickness)
x=tsin(i−r)cosr$x={\displaystyle \frac{tsin(i-r)}{cosr}}$
option D$D$ is correct 

Example

Calculate the Normal shift of a ray refracted through a rectangular glass slab

Problem: 
A glass slab of thickness 18 cm and refractive index 32${\displaystyle \frac{3}{2}}$ is placed on a printed matter. The normal shift of the printed matter is
Solution:

We know that, Refractive index is μ=Real DepthApparent Depth$\mu ={\displaystyle \frac{\text{Real Depth}}{\text{Apparent Depth}}}$
Given the Real depth dreal=18 cm$d_{\text{real}}=18\text{ cm}$Refractive index μ=1.5$\mu =1.5$
Thus, Apparent depth dapparent=drealμ=181.5=12 cm$d_{\text{apparent}}={\displaystyle \frac{d_{\text{real}}}{\mu }=\frac{18}{1.5}=12\text{ cm}}$
Thus Normal Shift = Real Depth - Apparent Depth=18−12=6 cm$=18-12=6\text{ cm}$

Result

Observe the refraction through a rectangular glass slab

Consider a rectangular glass slab ABCD as shown in fig. For refraction at AB
μ=sinisinr1$\mu ={\displaystyle \frac{sini}{sin{r}_1}}$...........(1)
for refraction at CD, 
1μ=sinr2sine${\displaystyle \frac{1}{\mu }}={\displaystyle \frac{sin{r}_2}{sine}}$..........(2)
But, r1=r2$r_1=r_2$, from eq. (1) and(2), we get
Angle of incidnece = Angle of emergence
i=e$i=e$

Diagram

Multiple image formation in a thick plane mirror

In case if the thick mirror being used, one may find that the incident ray and reflected ray do not meet at the same point on plane of separation of medium. This is because of the formation of multiple images due to multiple reflection.
Brightness in decreasing order: l′>l>l1>l2...$l^{\prime }>l>l_1>l_2...$

Example

Calculate lateral shift of a rectangular glass slab

Problem:
Refractive index of a rectangular glass slab is m=3√$m=\sqrt{3}$ . A light ray incident at an angle 60∘${60}^{\circ }$ is displaced laterally through 2.5 cm$2.5cm$. Find the distance traveled by light in the slab.

Solution: 

Given μ=3√$\mu =\sqrt{3}$, θa=600${\theta }_a={60}^0$
So, lateral shift: d=tsin(θa−θ′b)cosθ′b$d=t{\displaystyle \frac{\sin ({\theta }_a-{\theta }_b^{\prime })}{\cos {\theta }_b^{\prime }}}$ ...................(1)
Let the length travelled in glass be l$l$.
From geometry, t=lcosθ′b$t=l\cos {\theta }_b^{\prime }$ ..........................(2)
Substituting t$t$  from (2) in (1), we get:
 d=lsin(θa−θ′b)$d=lsin({\theta }_a-{\theta }_b^{\prime })$ .............................................(3)
From Snell's law, sinθasinθ′b=μ${\displaystyle \frac{\sin {\theta }_a}{\sin {\theta }_b^{\prime }}}=\mu$
⇒ sin600=3√sinθ′b$\Rightarrow \sin {60}^0=\sqrt{3}\sin {\theta }_b^{\prime }$
⇒ sinθ′b=12$\Rightarrow \sin {\theta }_b^{\prime }={\displaystyle \frac{1}{2}}$
⇒ θ′b=300$\Rightarrow {\theta }_b^{\prime }={30}^0$
Hence, from (3), we get:
2.5=lsin(600−300)$2.5=l\sin ({60}^0-{30}^0)$
⇒ l=5cm$\Rightarrow l=5cm$

BookMarks

Page 1  Page 2  Page 3  Page 4  Page 5  Page 6  Page 7  Page 8  Page 9  Page 10  Page 11
Page 12  Page 13  Page 14  Page 15  Page 16  Page 17  Page 18  Page 19  Page 20  Page 21 
Page 22  Page 23  Page 24