Question

A thin linear object of size 1 mm is kept along the principal-al axis of a convex lens of focal length 10 cm, the object is at 15 cm from the lens. The length of the image is
A. 1 mm
B. 6 mm
C. 2 mm
D. 8 mm

Answer 1

Hint: We are given focal length of convex lens. Convex lens is a converging lens and its focal length is positive. The object is placed at the left and taking sign convention it will be negative. We need to find the height of the image which can be found out using magnification.

Complete step by step answer:
Given the focal length of the lens is 10 cm. The focal length for a convex lens is always positive.
u= object distance= -15 cm
focal length of lens= f= 10 cm
using lens formula,
$
\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\
\implies \dfrac{1}{f}+\dfrac{1}{u}=\dfrac{1}{v} \\
\implies \dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{15} \\
\implies \dfrac{1}{v}=\dfrac{3-2}{30} \\
\implies \dfrac{1}{v}=\dfrac{1}{30} \\
\therefore v=30cm \\
$
So, the image is formed at a distance of 30 cm to the right. Now using formula for magnification, \[m=\dfrac{{{h}_{i}}}{{{h}_{0}}}=\dfrac{v}{u}\]
Height of the object= 1 mm = 0.001 m
$
\dfrac{{{h}_{i}}}{0.001}=\dfrac{30}{-15} \\
\implies {{h}_{i}}=-0.002m \\
\therefore {{h}_{i}}=-2mm \\
$

Here negative sign means the image is inverted, so the height of the image is 2mm. So, the correct option is (C).

Additional Information:
Power of a lens is defined as the reciprocal of the focal length in (m). It is measured in diaptore. Convex lens power is positive while for the concave lens it is negative. Power means the ability of the lens to bend to make a sharp image on the retina.

Note:
Magnification if greater than one then the image is enlarged and if it is less than one then the image is diminished. If the magnification is negative then image is inverted and if magnification is positive then image is erect. If magnification is unity then size of the image is equal to the size of the object.