Question

an object is placed at a distance of 30 from the pole of a concave lens of focal length 15cm. Find the position of the image. What is its magnification​

Answer 1

Answer:

the image is formed above principal axis . The image is formed on same side of the object. Explanation:Now since the object is placed on 30 cm . It means the object is placed at 2f because f=15cm

Answer 2

S O L U T I O N :

$\underline{\bf{Given\::}}$

• Focal length of lens, (f) = -15 cm .
• Distance of the object from lens, (u) = -30 cm.

$\underline{\bf{Explanation\::}}$

As we know that formula of the lens;

$\boxed{\bf{\dfrac{1}{f} =\dfrac{1}{v} -\dfrac{1}{u} }}$

A/q

$\mapsto\tt{\dfrac{1}{-15} =\dfrac{1}{v} -\dfrac{1}{(-30)} }$

$\mapsto\tt{\dfrac{1}{-15} =\dfrac{1}{v} +\dfrac{1}{30} }$

$\mapsto\tt{\dfrac{1}{v} =\dfrac{1}{-15} -\dfrac{1}{30} }$

$\mapsto\tt{\dfrac{1}{v} =\dfrac{-2 - 1}{30}}$

$\mapsto\tt{\dfrac{1}{v} =\dfrac{-3}{30}}$

$\mapsto\tt{30= - 3v \:\:\underbrace{\sf{cross-multiplication}}}$

$\mapsto\tt{v=-\cancel{30/3}}$

$\mapsto\bf{v=-10\:cm}$

∴ Image is formed at a distance of 10 cm from the lens .

Now,

As we know that magnification formula;

$\boxed{\bf{m=\frac{v}{-u} }}$

$\mapsto\tt{m=\cancel{\dfrac{-10}{-30} }}$

$\mapsto\tt{m=\dfrac{1}{3}}$

$\mapsto\bf{m=0.33\:cm}$

Thus;

The magnification will be 0.33 cm .