A convex lens of focal length f is placed at origin. Object is placed on negative x-axis anywhere between $- \infty$ to $+ \infty$ . Choose correct graph between 'u' and 'v'.

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Solution

The correct option is C
Using the lens formula,
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Here $f$ will remain constant, we can say that at
$u = - \infty$
$\Rightarrow v = f$
and as $u$ comes close to 0, $v$ becomes infinity.
This case is similar when object is coming from $+ \infty$ . The only difference being, $u = + v e a n d v = - v e$ .

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A convex lens of focal length f is placed at origin. Object is placed on negative x-axis anywhere between −∞ to +∞. Choose correct graph between 'u' and 'v'.

The correct option is C Using the lens formula, 1f=1v−1u Here f will remain constant, we can say that at u=−∞ ⇒v=f and as u comes close to 0, v becomes infinity. This case is similar when object is coming from +∞. The only difference being, u=+ve and v=−ve.

A convex lens of focal length f is placed at origin. Object is placed on negative x-axis anywhere between $- \infty$ to $+ \infty$ . Choose correct graph between 'u' and 'v'.