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Question:

An object is kept infront of a concave mirror of focal length 20cm. The image is 3 times the size of the object. Calculate two possible distances of the object from the mirror?

Answer:

Given that :

The focal length of the concave lens, f = - 20 cm

Magnification, m = 3

Two cases:

$begin mathsize 14px style bold First bold space bold case bold space bold when bold space bold the bold space bold image bold space bold is bold space bold virtual bold comma bold space bold m bold space bold equals bold 3 We space know space that space magnification comma space straight m equals minus straight v over straight u straight i. straight e. space minus straight v over straight u equals 3 straight v space equals minus 3 straight u According space to space mirror space formula : 1 over straight f equals 1 over straight u plus 1 over straight v fraction numerator 1 over denominator minus 20 end fraction equals 1 over straight u minus fraction numerator 1 over denominator 3 straight u end fraction fraction numerator 1 over denominator minus 20 end fraction equals fraction numerator 3 minus 1 over denominator 3 straight u end fraction fraction numerator 3 straight u over denominator 2 end fraction equals minus 20 3 straight u space equals minus 40 straight u space equals fraction numerator minus 40 over denominator 3 end fraction straight u equals minus 13.3 space cm Therefore space straight v space equals minus 3 cross times minus 13.3 space cm space equals 39.9 space cm The space distance space of space image space from space the space mirror space when space the space image space is space virtual space is space 39.9 space cm bold Second bold space bold case bold space bold when bold space bold the bold space bold image bold space bold is bold space bold real bold comma bold space bold m bold space bold equals bold minus bold 3 We space know space that space magnification comma space straight m equals minus straight v over straight u straight i. straight e. space minus straight v over straight u equals minus 3 straight v space equals 3 straight u According space to space mirror space formula : 1 over straight f equals 1 over straight u plus 1 over straight v fraction numerator 1 over denominator minus 20 end fraction equals 1 over straight u plus fraction numerator 1 over denominator 3 straight u end fraction fraction numerator 1 over denominator minus 20 end fraction equals fraction numerator 3 plus 1 over denominator 3 straight u end fraction fraction numerator 3 straight u over denominator 4 end fraction equals minus 20 3 straight u space equals minus 80 straight u space equals fraction numerator minus 80 over denominator 3 end fraction straight u equals minus 26.67 space cm Therefore space straight v space equals 3 cross times minus 26.67 space cm space equals minus 80 space cm The space distance space of space image space from space the space mirror space when space the space image space is space real space is space 80 space cm end style$

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