Optics Concept Page - 21

Example

Different refractive indices on the sides of a lens

Example: A thin convex lens made of glass of  μ=$\mu =$ 1.5 has refracting surfaces of radii of curvature 10 cm each The left space of lens contains air and right space is filled with water of refractive index 43${\displaystyle \frac{4}{3}}$. A parallel beam of light is incident on it. The position of image isSolution: μa=1,μg=1.5,μw=1.33,R1=10cm,R2=−10cm,u=∞${\mu }_a=1,{\mu }_g=1.5,{\mu }_w=1.33,R_1=10cm,R_2=-10cm,u=\mathrm{\infty }$Using the relation μwv+μau=μg−μaR1−μw−μgR2${\displaystyle \frac{{\mu }_w}{v}}+{\displaystyle \frac{{\mu }_a}{u}}={\displaystyle \frac{{\mu }_g-{\mu }_a}{R_1}}-{\displaystyle \frac{{\mu }_w-{\mu }_g}{R_2}}$
1.33v=1.5−110−1.33−1.510${\displaystyle \frac{1.33}{v}}={\displaystyle \frac{1.5-1}{10}}-{\displaystyle \frac{1.33-1.5}{10}}$
v=1.33×100.67=20cm$v={\displaystyle \frac{1.33\times 10}{0.67}}=20cm$

Example

Example using lens formula

An object is situated at a distance of 15cm from a convex lens of focal length 30 cm. The position of the image formed by it will be1v−1u=1f${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$
1v−1−15=130${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{-15}}={\displaystyle \frac{1}{30}}$
1v=130−115${\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{30}}-{\displaystyle \frac{1}{15}}$
1v=−130${\displaystyle \frac{1}{v}}={\displaystyle \frac{-1}{30}}$
v=−30$v=-30$

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 μ1${\mu }_1$ , so that its real image is formed at equidistant from sphere of radius R and refractive index μ2${\mu }_2$  which is also placed in the medium of refractive index μ1${\mu }_1$ as shown in figure?We know, μ1v−μ2u=μ2−μ1R${\displaystyle \frac{{\mu }_1}{v}}-{\displaystyle \frac{{\mu }_2}{u}}={\displaystyle \frac{{\mu }_2-{\mu }_1}{R}}$
For 1, we get:
μ2v−μ1−d=(μ2−μ1)R${\displaystyle \frac{{\mu }_2}{v}}-{\displaystyle \frac{{\mu }_1}{-d}}={\displaystyle \frac{({\mu }_2-{\mu }_1)}{R}}$ ,  −d=u=$-d=u=$ object distanceor, μ1v=μ2−μ1R−μ1d=t${\displaystyle \frac{{\mu }_1}{v}}={\displaystyle \frac{{\mu }_2-{\mu }_1}{R}}-{\displaystyle \frac{{\mu }_1}{d}}=t$For 2, we get:
μ1d−μ2(2R−v)=μ2−μ1R${\displaystyle \frac{{\mu }_1}{d}}-{\displaystyle \frac{{\mu }_2}{(2R-v)}}={\displaystyle \frac{{\mu }_2-{\mu }_1}{R}}$or, (μ2−μ1R−μ1d)=μ2(2R−v)=μ2v$({\displaystyle \frac{{\mu }_2-{\mu }_1}{R}}-{\displaystyle \frac{{\mu }_1}{d}})={\displaystyle \frac{{\mu }_2}{(2R-v)}}={\displaystyle \frac{{\mu }_2}{v}}$or, 2R−v=v$2R-v=v$or, R=v$R=v$So, μ1d=μ2−μ1r${\displaystyle \frac{{\mu }_1}{d}}={\displaystyle \frac{{\mu }_2-{\mu }_1}{r}}$or, d=(μ1μ2−μ1)R$d=\left({\displaystyle \frac{{\mu }_1}{{\mu }_2-{\mu }_1}}\right)R$

Formula

Refraction of light at spherical surfaces

Let,

n1$n_1$ - refractive index of medium from which rays incident.

n2$n_2$ - refractive index of another medium.

u - distance of object from pole of spherical surface

v - distance of image from pole of spherical surface

tanα=MNOM$tan\alpha ={\displaystyle \frac{MN}{OM}}$

tanγ=MNMC$tan\gamma ={\displaystyle \frac{MN}{MC}}$

tanβ=MNMI$tan\beta ={\displaystyle \frac{MN}{MI}}$

Now, for Δ$\mathrm{\Delta }$NOC, i is the exterior angle.

i=∠NOM+∠NCM$i=\mathrm{\angle }NOM+\mathrm{\angle }NCM$

i=MNOM+MNMC$i={\displaystyle \frac{MN}{OM}}+{\displaystyle \frac{MN}{MC}}$.........(1)

Similarly,

r=MNMC−MNMI$r={\displaystyle \frac{MN}{MC}}-{\displaystyle \frac{MN}{MI}}$.........(2)

Now by using snells law we get

n1 sin i=n2sin r$n_{1}sini=n_{2}sinr$

Or for small angles

n1 i=n2 r$n_{1}i=n_{2}r$

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

n1OM+n2MI=n2−n1MC${\displaystyle \frac{n_1}{OM}}+{\displaystyle \frac{n_2}{MI}}={\displaystyle \frac{n_2-n_1}{MC}}$

As,

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

Hence equation becomes

n2v−n1u=n2−n1R${\displaystyle \frac{n_2}{v}}-{\displaystyle \frac{n_1}{u}}={\displaystyle \frac{n_2-n_1}{R}}$

Example

Calculate Focal Length of Lenses using lens maker's formula

Example: A plano-convex lens has thickness 4 cm$4cm$. 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 cm$3cm$. 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 258 cm${\displaystyle \frac{25}{8}}cm$. Find the focal length of the lens.

Solution:
When the curved surface of the lens (refractive index μ$\mu$) 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 I1$I_1$.
Here, u1=AO=−4 cm,v1=AI1=3 cm$u_1=AO=-4cm,v_1=A{I}_1=3cm$, μ1=μ${\mu }_1=\mu$, μ2=1${\mu }_2=1$, and R1=∞$R_1=\mathrm{\infty }$
∴ μ2v1−μ1u1=μ2−μ1R1$\therefore {\displaystyle \frac{{\mu }_2}{v_1}}-{\displaystyle \frac{{\mu }_1}{u_1}}={\displaystyle \frac{{\mu }_2-{\mu }_1}{R_1}}$ gives, 
1−3−μ−4=1−μ∞${\displaystyle \frac{1}{-3}}-{\displaystyle \frac{\mu }{-4}}={\displaystyle \frac{1-\mu }{\mathrm{\infty }}}$................ (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 I2$I_2$.
Here, u2=AO=−4 cm,v2=AI2=−258 cm$u_2=AO=-4cm,v_2=A{I}_2=-{\displaystyle \frac{25}{8}}cm$, μ1=μ,μ2=1${\mu }_1=\mu ,{\mu }_2=1$, and R2=−R$R_2=-R$
∴μ2v2−μ1u2=μ2−μ1R2$\therefore {\displaystyle \frac{{\mu }_2}{v_2}}-{\displaystyle \frac{{\mu }_1}{u_2}}={\displaystyle \frac{{\mu }_2-{\mu }_1}{R_2}}$ gives.
1(−258)−μ−4=1−μ−R${\displaystyle \frac{1}{(-{\displaystyle \frac{25}{8}})}}-{\displaystyle \frac{\mu }{-4}}={\displaystyle \frac{1-\mu }{-R}}$
From Eq. (i), we get:
μ=4/3$\mu =4/3$, therefore this equation gives
−825+4/34=−(1−43)R=−825+13=13R$-{\displaystyle \frac{8}{25}}+{\displaystyle \frac{4/3}{4}}=-{\displaystyle \frac{(1-{\displaystyle \frac{4}{3}})}{R}}=-{\displaystyle \frac{8}{25}}+{\displaystyle \frac{1}{3}}={\displaystyle \frac{1}{3R}}$
or 175=13R${\displaystyle \frac{1}{75}}={\displaystyle \frac{1}{3R}}$
This gives R=25 cm$R=25cm$
The focal length (f$f$) of plano-convex lens (R1=R$(R_1=R$ and R2=∞)$R_2=\mathrm{\infty })$ is 1f=(μ−1)(1R−1∞)=μ−1R=43−125=175${\displaystyle \frac{1}{f}}=(\mu -1)({\displaystyle \frac{1}{R}}-{\displaystyle \frac{1}{\mathrm{\infty }}})={\displaystyle \frac{\mu -1}{R}}={\displaystyle \frac{{\displaystyle \frac{4}{3}}-1}{25}}={\displaystyle \frac{1}{75}}$
⇒ f=75 cm$\Rightarrow f=75cm$

Example

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

Distance between object and image =2f+2f=4f

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