If the image of an object is at the focal point (f) to the light side of a convex lens, the position of the object on the left of the lens is at

Question

If the image of an object is at the focal point (f) to the light side of a convex lens, the position of the object on the left of the lens is at (A) f (B) 2 f (C) ∞ (D) none of the above

Tardigrade Admin · Accepted Answer

The given situation is shown in the following figure

Given,

v = + f, f = + f

By

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

We have,

\frac{1}{f} - \frac{1}{f} = \frac{1}{u}

\Rightarrow \frac{1}{u} = 0

\Rightarrow u = \infty

So, object is at

\infty

distance on right side of lens.

Q. If the image of an object is at the focal point (f) to the light side of a convex lens, the position of the object on the left of the lens is at

f

2 f

∞

none of the above

The given situation is shown in the following figure Given, v=+f,f=+f By v1​−u1​=f1​ We have, f1​−f1​=u1​ ⇒u1​=0 ⇒u=∞ So, object is at ∞ distance on right side of lens.

Q. If the image of an object is at the focal point $(f)$ to the light side of a convex lens, the position of the object on the left of the lens is at

$\infty$

The given situation is shown in the following figure

Given, $v = + f, f = + f$
By $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
We have, $\frac{1}{f} - \frac{1}{f} = \frac{1}{u}$
$\Rightarrow \frac{1}{u} = 0$
$\Rightarrow u = \infty$
So, object is at $\infty$ distance on right side of lens.