Question

An object is placed at a distance of 10cm from a concave lens of focal length 20cm.Find the position of the image and discuss nature.​

Answer 1

Answer:

Given,

object distance, u = -10 cm

Focal length  , f = +20 cm

we have lens formula,

         $\frac{1}{v} -\frac{1}{u} = \frac{1}{f} \\\\by \ putting \ the\ value \ we \ get \ ,\\\\=>\frac{1}{v} - \frac{1}{-10} = \frac{1}{20}\\\\=> \frac{1}{v} = \frac{1}{20} - \frac{1}{10}\\\\=> v = \frac{200}{-10} = -20 \ cm \\\\That \ means \ that \ image\ is \ virtual \ and \ erect \ and \ formed \ in \ negative \ y-axis\ , \ left\ of\ the\ lens\ \ and \ between \ the \ focal length\ and\ the\ optical\ centre\ at\ left \ side(Ans.)$  

       Hope it helps

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Answer 2

Explanation:

1f=1u−1v, called the lens formula where 'u' is the object distance and 'v' is the image distance measured from the pole of the lens of focal length 'f'. Hence the image will be formed on the left side of the lens (shown by negative sign) at a distance of 20 cm.

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