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Subjects
Science CBSE
Class 10
Light reflection and refraction
Textbook questions

7. Exercises VII

Exercise condition:

5 m.

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.

Lens formula is,

\(\frac{1}{u}\ =\ \frac{1}{v}\ +\ \frac{1}{f}\)
\(\frac{1}{v}\ -\ \frac{1}{u}\ =\ \frac{1}{f}\)
\(\frac{1}{u}\ -\ \frac{1}{v}\ =\ \frac{1}{f}\)
\(\frac{1}{v}\ =\ \frac{1}{f}\ -\ \frac{1}{u}\)

For converging lens,

\(M\ =\ \frac{h_i}{h_o}\ =\ \frac{v}{u}\)
\(M\ =\ \frac{h_i}{h_o}\ =\ \frac{u}{v}\)
\(M\ =\ \frac{h_o}{h_i}\ =\ \frac{u}{v}\)
\(M\ =\ \frac{h_o}{h_i}\ =\ \frac{v}{u}\)

Image distance (\(v\)) \(=\) \(cm\)

Height of the image (\(h_i\)) \(=\) \(cm\)

Nature of the image is,

real, inverted, and same size
virtual, erect and diminished size
real, inverted and diminished size
virtual, erect and same size

Important!

Note: Enter your answers with two decimal digits. Answer points are provided only for formulas and numerical answers. Draw the ray diagram in the white and compare it with the given answer.

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