An object of height $h_{0} = 1 cm$ is moved along principal axis of a convex lens of focal length $f = 10 cm$ .Figure shows variation of magnitude of height of image with image distance (v). Find $v_{2} - v_{1}$ in $c m$ _______.

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Solution

We know that, $\frac{h_{i}}{h_{0}} = \frac{f - v}{f} \Rightarrow h_{i} = - (\frac{v}{f}) h_{0} + h_{0}$

$\Rightarrow | h_{i} | = - \frac{v}{f} h_{0} + h_{0} (- \infty \leq v \leq f)$
or $| h_{i} | = \frac{v}{f} h_{0} - h_{0} (f \leq v \leq - \infty) (1)$

So, from figure, $h_{2} = h_{0} = 1 cm$

Therefore, $v_{2} = 2 f$

For $v_{1}$ , using $(1)$ we get, $v_{1} = f$

$∴ v_{2} - v_{1} = f = 10 cm$

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An object of height h0=1 cm is moved along principal axis of a convex lens of focal length f=10 cm.Figure shows variation of magnitude of height of image with image distance (v). Find v2−v1 in cm_______.

We know that,hih0=f−vf⇒hi=−(vf)h0+h0 ⇒|hi|=−vfh0+h0 (−∞≤v≤f) or |hi|=vfh0−h0 (f≤v≤−∞) (1) So, from figure, h2=h0=1 cm Therefore, v2=2f For v1 , using (1) we get, v1=f ∴v2−v1=f=10 cm

An object of height $h_{0} = 1 cm$ is moved along principal axis of a convex lens of focal length $f = 10 cm$ .Figure shows variation of magnitude of height of image with image distance (v). Find $v_{2} - v_{1}$ in $c m$ _______.