Question

✴️✳️ solve .....

an object is placed at a distance of 40cm from a concave lens of focal length 20cm . find the nature and position of the image ?


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Answer 1

solution

in question we find position of the image which is given by the I'mage distance v.

Here given in question,

Objects distance u = - 40cm ( it is to left of lens)

I'mage distance v = ?

Focal length f = - 20 cm ( concave lens )

here using the lens formula,

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

given values put in the lens formula .

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

$\frac{1}{-40} - \frac{1}{u} = \frac{1}{-20}$

$\frac{1}{v} - \frac{1}{40} = \frac{-1}{20}$

$\frac{1}{v}$ =$\frac{-1}{20}$ - $\frac{1}{40}$

$\frac{1}{v}$ =$\frac{-20- 1}{40}$

$\frac{1}{v}$ = $\frac{-21}{40}$

v = $\frac{40}{-21}$

v = -1.90 cm

since the I'mage distance v = -1.90

so , the I'mage formation at a distance of 1.90 cm from the concave lens .

and the minus sign for I'mage distance shows that the because I'mage formation left side of the concave lens.

and we also know that the property of concave lens is always form a virtual and erect I'mage , .

this cause nature of I'mage is virtual and erect.

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