8. TBQ - Magnification

Question:

5 m.

Calculate all the values in the given question:

An object \(4\ cm\) in length is held \(50\ cm\) away from a converging lens of focal length \(12\ cm\). Find the position, size and the nature of the image formed.

Lens formula is,

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

For converging lens,

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

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

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

Nature of the image is,

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

[Note: Enter your answers with two decimal digits.]

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