Locate the virtual image of an object 2 m in front of the eye glasses.

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Solution

When the object object distance is infinite ,A virtual image has to be formed at far point.
$u = \infty$ and $v = - 5 m$
So $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{f} = \frac{1}{- 5} - \frac{1}{\infty}$
So $f = - 5 m$
Now when object is placed at 2m , so $u = - 2 m$ and $f = - 5$
So $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{- 5} = \frac{1}{v} - \frac{1}{- 2}$
$v = - 1.43 m$

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Locate the virtual image of an object 2 m in front of the eye glasses.

When the object object distance is infinite ,A virtual image has to be formed at far point.u=∞ and v=−5mSo 1f=1v−1u 1f=1−5−1∞ So f=−5mNow when object is placed at 2m , so u=−2m and f=−5So 1f=1v−1u 1−5=1v−1−2 v=−1.43m