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Numerical · Q27

Q.An object 5 cm5\ \text{cm} tall is placed 30 cm30\ \text{cm} in front of a convex mirror of focal length 20 cm20\ \text{cm}. Find the position, size and nature of the image formed.

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For the convex mirror, f=+20 cmf=+20\ \text{cm}, u=−30 cmu=-30\ \text{cm}. From the mirror formula, 1v=1f−1u=120−1−30=120+130=3+260=560=112\dfrac1v=\dfrac1f-\dfrac1u=\dfrac1{20}-\dfrac1{-30}=\dfrac1{20}+\dfrac1{30}=\dfrac{3+2}{60}=\dfrac5{60}=\dfrac1{12}, so v=+12 cmv=+12\ \text{cm} (virtual, behind the mirror). The magnification is m=−v/u=−12−30=+0.4m=-v/u=-\dfrac{12}{-30}=+0.4, so the image height is hi=m×ho=0.4×5 cm=2 cmh_i=m\times h_o=0.4\times5\ \text{cm}=2\ \text{cm}. The positive mm confirms the image is erect, and ∣m∣<1|m|<1 confirms it is diminished — exactly the behaviour a convex mirror always gives, whatever the object distance.

✓Final answer

v=+12 cmv=+12\ \text{cm} (virtual, behind the mirror), image height 2 cm2\ \text{cm}, erect and diminished.

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