# A converging lens of 2.5 cm focal length is used as a simple microscope producing virtual image at 25 cm from the eye. The position of the object from the lens is (nearly) 2 cm.

## Question

A converging lens of 2.5 cm focal length is used as a simple microscope producing virtual image at 25 cm from the eye. The position of the object from the lens is (nearly) 2 cm.

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## Answer

To solve this problem, we need to find the object distance ($ u $) given the focal length ($ f $) of a converging lens and the image distance ($ v $).

Given:

- Focal length, $ f = 2.5 $ cm
- Image distance, $ v = 25 $ cm (since the image is virtual, it will be negative in the lens formula, i.e., $ v = -25 $ cm)

We use the lens formula for a converging lens: $$ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} $$

Rearranging to solve for $ \frac{1}{u} $: $$ \frac{1}{u} = \frac{1}{v} - \frac{1}{f} $$

Substituting the given values: $$ \frac{1}{u} = \frac{1}{-25} - \frac{1}{2.5} $$

Converting $ 2.5 $ to a fraction for easier calculation: $$ \frac{1}{2.5} = \frac{1}{\frac{25}{10}} = \frac{10}{25} $$

So, $$ \frac{1}{u} = \frac{1}{-25} - \frac{10}{25} = \frac{1}{-25} - \frac{10}{25} = -\frac{1}{25} - \frac{10}{25} = -\frac{11}{25} $$

Taking the reciprocal to find $ u $: $$ u = -\frac{25}{11} \approx -2.27 \text{ cm} $$

Since the object distance comes out negative, it indicates that the **object is placed on the left side of the lens**.

Therefore, the position of the object from the lens is **nearly 2 cm** (specifically, about 2.27 cm for more precise calculations).

**Answer:** The object distance from the lens is nearly **2 cm**.

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