Optics Concept Page - 20

Formula

Calculate Newton's formula for thick lenses

If the distance of object and image are not measured from optical centre (C), but from first and second principal foci respectively, then newtons formula states
f1f2=x1x2$f_1{f}_2=x_1{x}_2$
where,
x1$x_1$ = distance of object from first focus F1$F_1$.
x2$x_2$ = distance of image from second focus F2$F_2$
If medium on either side of lens is same, then
−f1=f2=f$-f_1=f_2=f$
So, Newtons formula takes the form,x1x2=−f2$x_1{x}_2=-f^2$

Example

Image formation from compound lenses

A converging lens of focal length 30 cm$30cm$ and diverging lens of focal length 20 cm$20cm$ are kept 15 cm$15cm$ apart with their principal axes coinciding. Where shall an object be placed to form an image at infinity?
Two cases are possible.
Case 1:
Final image is formed by the concave lens.
For concave lens v=∞$v=\mathrm{\infty }$; f=−20 cm$f=-20cm$
⇒$\Rightarrow$ u=+20 cm$u=+20cm$
Now, v=20+15=35 cm$v=20+15=35cm$ serves as image distance for the convex lens.
So, v=+35 cm$v=+35cm$; f=+30 cm$f=+30cm$
1v−1u=1f${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$
⇒$\Rightarrow$ 135−1u=130${\displaystyle \frac{1}{35}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{30}}$
⇒$\Rightarrow$ u=−210 cm=210 cm$u=-210cm=210cm$ from converging lens.

Case 2:
Final image is formed by convex lens. 
So,v=∞$v=\mathrm{\infty }$; f=+30 cm$f=+30cm$
⇒$\Rightarrow$ u=−30 cm$u=-30cm$ for convex lens to form image at infinity 
So for the concave lens, the image distance is v=−30−(−15)=−15 cm$v=-30-(-15)=-15cm$; f=−20 cm$f=-20cm$ 
1v−1u=1f${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$
⇒$\Rightarrow$ 1−15−1u=1−20${\displaystyle \frac{1}{-15}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{-20}}$
or, u=−60 cm$u=-60cm$, i.e, 60 cm$60cm$ from diverging lens.

Example

Formation of image from a combination of lenses, mirrors and glass slabs

Example: A convex lens of focal length 40 cm$40cm$ is held at a distance 12 cm$12cm$ co-axially above a concave mirror of focal length 18 cm$18cm$. If the convex lens is replaced by a glass plate of thickness 6 cm$6cm$, refractive index μ=32$\mu ={\displaystyle \frac{3}{2}}$ and gives rise to an image coincident with itself, then what will be the value of d$d$?

Solution:
When the  ray from O passes through the slab of refractive index  (μ$\mu$), then there will be shift of point O to I1$I_1$ and then this point I1$I_1$ will act as source for the concave mirror.
Shift = (1−1μ)t$(1-{\displaystyle \frac{1}{\mu }})t$=(1−23)6$=(1-{\displaystyle \frac{2}{3}})6$
⟹$⟹$ shift =2 cm$=2cm$, i.e., the object will appear to look closer by 2 cm$2cm$.
Now as the final image is formed at a point O itself, so the ray from point I1$I_1$ will retrace its own path (i-e, I1$I_1$ should be at R$R$ of concave mirror).
So, d−2+12=2×f1=2×18=36$d-2+12=2\times f_1=2\times 18=36$
⟹ d=26 cm$⟹d=26cm$ 

Example

Image formation by broken lenses

A point object O is placed at a distance 30 cm$30cm$ from a convex lens (focal length 20 cm$20cm$) cut into two halves each of which is displaced by 0.05 cm$0.05cm$ perpendicular to the principal axis. What is the distance between the two images formed?
1v−1u=1f${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$
or, 1v−1−30=−120${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{-30}}={\displaystyle \frac{-1}{20}}$
or, v=60cm$v=60cm$
So, d=3×h=0.1×3=0.3 cm$d=3\times h=0.1\times 3=0.3cm$
m=vu=hiho$m={\displaystyle \frac{v}{u}}={\displaystyle \frac{h_i}{h_o}}$=6030=2=hi0.05$={\displaystyle \frac{60}{30}}=2={\displaystyle \frac{h_i}{0.05}}$
or, hi=0.1 cm$h_i=0.1cm$

Example

Image formation for lenses with one side silvered

A plano convex lens of focal length 30 cm$30cm$ has its plane surface silvered. An object is placed 40 cm$40cm$ from the lens on the convex side. The distance of the image from the lens is:1v−1u=1f$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
or, 1v=130−140$\frac{1}{v}=\frac{1}{30}-\frac{1}{40}$
or, 1v=4−3120$\frac{1}{v}=\frac{4-3}{120}$
or, v=+120 cm$v=+120cm$
Now, this image acts as the object for the lens.So, u=+120 cm, f=30 cm$u=+120cm,f=30cm$
1v−1u=1f$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
or, 1v=130+1120$\frac{1}{v}=\frac{1}{30}+\frac{1}{120}$
or, v=+24 cm$v=+24cm$

Example

Image formation from lenses where object is being moved

Example: A thin converging lens of focal length f=25 cm$f=25cm$ forms the image of an object on a screen placed at a distance of 75 cm$75cm$ from the lens. The screen is moved closer to the lens by a distance of 25 cm$25cm$. Find the distance through which the object has to be shifted so that its image on the screen is sharp again.

Solution:
Focal length of the converging lens: f=25 cm$f=25cm$
Case 1: Image distance: v=75 cm$v=75cm$.
Let object distance be u$u$. 
Using 1v−1u=1f${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$, we get:
175−1u=125${\displaystyle \frac{1}{75}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{25}}$or, u=37.5 cm$u=37.5cm$This means that the object is at a distance of 37.5 cm$37.5cm$ from the lens initially.
Case 2: The screen is shifted by 25 cm$25cm$ towards the lens.
New image distance: v=75−25=50 cm$v=75-25=50cm$Let new object distance be u$u$Using: 1v−1u=1f${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{f}}$, we get:
150−1u=125${\displaystyle \frac{1}{50}}-{\displaystyle \frac{1}{u}}={\displaystyle \frac{1}{25}}$or, u=50 cm$u=50cm$ So, the object has to be shifted by a distance: d=50−37.5=12.5 cm$d=50-37.5=12.5cm$ away from the lens.

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