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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