Question

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Answer 1

Object distance, \(u = −25\ cm\)
Object height, \(h_o = 5\ cm\)
Focal length, \(f = +10\ cm\)
Using the lens formula:
\(\frac 1v-\frac1u=\frac 1f\)

\(\frac 1v=\frac 1f+\frac 1u\)

\(\frac 1v=\frac {1}{10} -\frac {1}{25}\)

\(\frac 1v=\frac {15}{250}\)

\(v=\frac {250}{15}\)
\(v=16.66\ cm\)
The positive image distance indicates the image is formed on the opposite side of the lens.
Magnification, \(m=-\frac {\text {Image\ distance}}{\text {Object\ distance}}\)
\(m =-\frac vu\)

\(m =-\frac {16.66}{25}\)
\(m =-0.66\)
The negative magnification signifies a real image formed behind the lens.
Magnification, \(m=-\frac {\text {Image\ height}}{\text {Object\ height}}\)
\(m=\frac {H_I}{H_o}\)

\(m=\frac {HI}{5}\)
\(HI=5 \times m\)
\(HI =5 \times (-0.66)\)
\(HI=-3.3\ cm\)
The negative image height indicates the image formed is inverted.
The ray diagram below illustrates the image's position, size, and nature.