Question.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object ? Also, find the power of the lens.
A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object ? Also, find the power of the lens.
solution:
Given, $\mathrm{v}=+50 \mathrm{~cm}$ (positive sign is taken because real image in convex lens is formed on the right side).
Since, the image is real and size of the image is equal to the size of object.
$\therefore$ Magnification, $\mathrm{m}=-1$
Now, $m=+\frac{v}{u}$ or $-1=+\frac{v}{u}$
or $u=-v=-(+50)=-50 \mathrm{~cm}$
By lens equation,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{(+50)}-\frac{1}{(-50)}=\frac{1}{f}$
$\frac{1}{f}=\frac{2}{50}=\frac{1}{25} \quad$ or $f=+25 \mathrm{~cm}=+0.25 \mathrm{~m}$
Power, $P=\frac{1}{f}=\frac{1}{+0.25}=+4$ Dioptre
Given, $\mathrm{v}=+50 \mathrm{~cm}$ (positive sign is taken because real image in convex lens is formed on the right side).
Since, the image is real and size of the image is equal to the size of object.
$\therefore$ Magnification, $\mathrm{m}=-1$
Now, $m=+\frac{v}{u}$ or $-1=+\frac{v}{u}$
or $u=-v=-(+50)=-50 \mathrm{~cm}$
By lens equation,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\frac{1}{(+50)}-\frac{1}{(-50)}=\frac{1}{f}$
$\frac{1}{f}=\frac{2}{50}=\frac{1}{25} \quad$ or $f=+25 \mathrm{~cm}=+0.25 \mathrm{~m}$
Power, $P=\frac{1}{f}=\frac{1}{+0.25}=+4$ Dioptre