Question

Virtual image of an object is placed between P and F of a concave mirror is larger than the object.What is magnification in this case?​

Answer 1

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Answer 2

Given:

A object is placed between the pole and the focus of a concave mirror resulting in formation of a larger virtual image.

To find:

Magnification in this case?

Calculation:

In case of mirrors , if we consider the object to be erect then virtual image of that same object will always be erect.

Hence, the sign of magnification will always be positive.

Now , whenever we place an object between the pole and focus of a concave mirror we get an image characteristics as follows:

• Image is virtual.
• Image is erect.
• Image is magnified.

$\boxed{ \bold{ \therefore \: magnification = \dfrac{h_{i}}{h_{o} } }}$

• Here , the magnification value is greater than zero (which signifies erect image) , but lesser than 1 (which signifies diminished image).

Refer to the diagram to understand the image characteristics.

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