Question

an object 8.25 cm from a lens creates a virtual image of magnification 2.40 what is the focal length of the lens (mind your minus sign) (unit=cm)

Answer 1

Answer:

Focal length of the lens is 14.14 cm

Explanation:

As we know that the image formed due to lens is virtual image

so here we have

M = 2.40

now we have

$\frac{d_i}{d_o} = 2.40$

now we have

distance of object is 8.25 cm

so we have

$d_i = 8.25(2.40) = 19.8 cm$

now by lens formula

$\frac{1}{d_i} - \frac{1}{d_o} = \frac{1}{f}$

$\frac{1}{-19.8} - \frac{1}{-8.25} = \frac{1}{f}$

$f = 14.14 cm$

Answer 2

The focal length of the lens is 14.1428 cm and it is a positive focal length. The lens is convex.

Given information:

The object distance is $u=-8.25$ cm.

The magnification of the lens is $m=2.4$

So, the image distance will be calculated as,

$m=\dfrac{v}{u}\\2.4=\dfrac{v}{-8.25}\\v=-19.8\rm\; cm$

Now, use the lens formula to calculate the focal length f of the lens.

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\dfrac{1}{f}=\dfrac{1}{-19.8}-\dfrac{1}{-8.25}\\\dfrac{1}{f}=0.070\\f=+14.1428$

Therefore, the focal length of the lens is 14.1428 cm and it is a positive focal length. The lens is convex.

For more details, refer to the link:

https://brainly.com/question/766997