Question

At what distance should an object be placed from a convex lens of focal length 18 cm to obtain an image at 24 cm from it on the other side. What will be the magnification produced in this case?


Hoping for a well answer!
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Answer 1

Answer:

The focal length of a convex lens, f = 18 cm.

Image distance, v = 24 cm

Object distance, u = ?

To find- Magnification

Solution:

By using lens formula-

where, v = image distance, u = object distance, and f = focal length

Substituting the values of f, v and u we get,

So, the object distance is -72cm

The object should be placed at a distance of -72 cm from the lens.

Now, the equation for finding magnification of a lens can be given as-

Substituting the values in magnification formula we get-

Hence, the magnification produced will be

Answer 2

$\bull$ Qυєѕτiοи :-

: At what distance should an object be placed from a convex lens of focal length 18 cm to obtain an image at 24 cm from it on the other side. What will be the magnification produced in this case?

$\bull$ Given :-

f = 18 cm

v = 24 cm

$\bull$ To Find :-

Magnification produced in this case

$\bull$ Formula Used :-

In this question we will use the lens formula and magnification formula.

✪ Lens Formula :-

$\bigstar \: \boxed{ \sf{ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} }} \: \bigstar$

Where,

• f = focal length
• u = object distance
• v = image distance

✪ Magnification Formula :-

$\bigstar \: \boxed{ \sf \: m = \frac{v}{u} } \: \bigstar$

Where,

• m = magnification
• v = image distance
• u = object distance

$\bull$ Solution :-

✪ First of all we will use, lens formula to find out the value of u i.e. object distance. Therefore,

$\sf : \implies \: \dfrac{1}{18} = \dfrac{1}{24} - \dfrac{1}{u} \\ \\ \sf : \implies \: \dfrac{1}{u} = \dfrac{1}{24} - \dfrac{1}{18} \\ \\ \sf : \implies \: \dfrac{1}{u} = - \dfrac{1}{72} \: \: \: \: \: \: \:$

✪ Now, we will find out the magnification using magnification formula. Therefore,

$\sf : \implies \: m = \dfrac{24}{72} \\ \\ \sf : \implies \: m = \dfrac{1}{3}$

★ Therefore, magnification produced is this case is $\sf \dfrac{1}{3}$