Optics Concept Page - 22

Example

Minimum distance between object and its real image for a convex lens

Example: In a convex lens of focal length f$f$, find the minimum distance between an object and its real image.

Solution:
Let image distance be v$v$, object distance be −u$-u$ and focal length be f$f$.
Let distance between object and image be s$s$.
∴s=u+v$\therefore s=u+v$
Now, 1v+1u=1f$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
or, u+vuv=1f$\frac{u+v}{uv}=\frac{1}{f}$or, su(s−u)=1f$\frac{s}{u(s-u)}=\frac{1}{f}$
or, s=u2u−f$s=\frac{u^2}{u-f}$
For point of minima, dsdu=0$\frac{ds}{du}=0$
or, u2−2uf(u−f)2=0$\frac{u^2-2uf}{(u-f{)}^2}=0$
or, u=0,2f$u=0,2f$
Therefore, s=(2f)22f−f=4f$s=\frac{(2f{)}^2}{2f-f}=4f$

Definition

Magnification

Magnification of a lens is defined as:
m=IO=vu$m={\displaystyle \frac{I}{O}}={\displaystyle \frac{v}{u}}$
Note:
Sign convention must be followed while using formula for magnification. Hence, it can be positive or negative.
|m|>1⟹$|m|>1⟹$ image is magnified.
|m|=1⟹$|m|=1⟹$ image is same size as object.
|m|<1⟹$|m|<1⟹$ image is diminished.

Example

Calculate magnification of lens

Problem:
The focal length of a thin biconvex lens is 20cm$20cm$. When an object is moved from a distance of 25cm$25cm$ in front of it to 50cm$50cm$, the magnification of its image changes from m25$m_{25}$ to m50$m_{50}$. The ratio m25m50${\displaystyle \frac{m_{25}}{m_{50}}}$ is : 
Solution:
When object is at 25cm$25cm$

u=−25cm$u=-25cm$, v=100cm$v=100cm$
m25=vu=10025$m_{25}={\displaystyle \frac{v}{u}}={\displaystyle \frac{100}{25}}$.....(1)
When object is at 50cm$50cm$
u=−50cm$u=-50cm$, v=100/3cm$v=100/3cm$
m50=vu=100150$m_{50}={\displaystyle \frac{v}{u}}={\displaystyle \frac{100}{150}}$.....(2)
We have to find m25m50${\displaystyle \frac{m_{25}}{m_{50}}}$
m25m50=10025100150=6${\displaystyle \frac{m_{25}}{m_{50}}}={\displaystyle \frac{{\displaystyle \frac{100}{25}}}{{\displaystyle \frac{100}{150}}}}=6$

Example

Calculate object distance, image distance and focal length of lens

Problem: 1
An object is placed at a distance of 10 cm from a convex lens of focal length 12 cm .Find the position and nature of the image.
Solution:
Given, u=$=$-10cm
f=$=$12cm
v=$=$?
we know
1f=1v−1u${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}$
112=1v−(−1)10${\displaystyle \frac{1}{12}}={\displaystyle \frac{1}{v}}-{\displaystyle \frac{(-1)}{10}}$
1v=5−660${\displaystyle \frac{1}{v}}={\displaystyle \frac{5-6}{60}}$
v=−60 cm$v=-60cm$
Hence, image formed behind the mirror at a distance 60 cm and the image is virtual and magnified beyond 2F.
Problem:2
A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 24 cm. The distance of the object from the lens is 16 cm. Find the position, size and nature of the image formed, using the lens formula.
Solution:Given  -    f=+24cm,u=−16cm$f=+24cm,u=-16cm$ ,As |u|(16cm)<|f|(24cm)$|u|(16cm)<|f|(24cm)$ ,it means object lies between F$F$ and C$C$ , in this position of object the rays from object cannot meet at any point on the other side of lens ,which is also clear from the position of image found below ,From lens equation  v=ufu+f=−16×24−16+24=−48cm$v={\displaystyle \frac{uf}{u+f}}={\displaystyle \frac{-16\times 24}{-16+24}}=-48cm$ ,-ive sign shows that image will be formed on the same side of lens , where the object is placed .
Now linear magnification     m=I/O=v/u=−48/−16=+3$m=I/O=v/u=-48/-16=+3$ ,or                                        I=3×4=12cm$I=3\times 4=12cm$       (given  O=4cm$O=4cm$) ,since m=+ive$m=+ive$ and m>1$m>1$ ,therefore image will be virtual ,erect and magnified .

Example

combination of lenses

Problem:
A convex lens, of focal length 30cm$30cm$, a concave lens of focal length 120cm$120cm$, and a plane mirror are arranged as shown. For an object kept at a distance of 60cm$60cm$ from the convex lens, the final image, formed by the combination, is a real image, at a distance of :
Solution:
Location of image formation after first refraction from the lens,1v+160=130${\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{60}}={\displaystyle \frac{1}{30}}$.
This gives v=60cm$v=60cm$
Next refraction from concave lens, 1v2−140=−1120${\displaystyle \frac{1}{v_2}}-{\displaystyle \frac{1}{40}}=-{\displaystyle \frac{1}{120}}$
Hence v2=60cm$v_2=60cm$
Since this will be formed behind the mirror, this will be final image.It is at a distance of 60cm$60cm$ from the concave lens.

Example

Image formation when multiple lenses are used

What should be the value of distance d$d$ so that final image is formed on the object itself? (Focal lengths of the lenses are written on the lenses.)Image after refraction from first lens: 1v1−1−10=110${\displaystyle \frac{1}{v_1}}-{\displaystyle \frac{1}{-10}}={\displaystyle \frac{1}{10}}$
or, v1=∞$v_1=\mathrm{\infty }$
Image after second refraction from concave lens, 1v−1∞=1−20${\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{\mathrm{\infty }}}={\displaystyle \frac{1}{-20}}$
Hence, v=−20 cm$v=-20cm$
or, |v|=d+10 cm$|v|=d+10cm$ (If final image is formed on object itself)
Hence, d=10 cm

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