Subjects
Physics CBSE Live product
Pre-Crash Course
Light reflection and refraction

4. TBQ - Mirror formula and magnification

Question:

5 m.

Solve the problem below:

An object is placed at a distance of \(10\ cm\) from a convex mirror of focal length \(15\ cm\). Find the image position and its nature.

Mirror formula is,

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

Magnification is,

\(M\ =\ \frac{u}{v}\)
\(M\ =\ \frac{-v}{u}\)
\(M\ =\ \frac{v}{u}\)
\(M\ =\ \frac{-u}{v}\)

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

Magnification (\(M\)) \(=\)

Nature of the image is,

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

[Note: Enter your answers with one decimal digit.]

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