An object is placed at a distance of f 2 from a convex lens. The image will be Options: A . &nbspAt one of the foci, virtual and double its size B . &nbspAt 3f2, real and inverted C . &nbspAt 2f, virtual and erect D . &nbspNone of these

Answer: Option A : A 1 f = 1 v − 1 u (Given u = − f 2 ) ⇒ 1 f = 1 v + ( 1 f 2 ) ⇒ 1 v = 1 f − 2 f ⇒ 1 v = − 1 f and m = v u = f f 2 = 2 So virtual at the focus and of double size.

An object is placed at a distance of f2 from a convex lens. The image will

Question

An object is placed at a distance of

\frac{f}{2}

from a convex lens. The image will be

Options:

A . At one of the foci, virtual and double its size

B . At 3f2, real and inverted

C . At 2f, virtual and erect

D . None of these

Answer: Option A
:
A

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

(Given

u = \frac{- f}{2}

)

\Rightarrow \frac{1}{f} = \frac{1}{v} + (\frac{1}{\frac{f}{2}}) \Rightarrow \frac{1}{v} = \frac{1}{f} - \frac{2}{f}

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

and

m = \frac{v}{u} = \frac{f}{\frac{f}{2}}

= 2
So virtual at the focus and of double size.

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