Question

2. The object distance of a lens is -30cm
and image distance is -10cm. Find the
magnification of the lens. With the help
of this, decide whether the size of the
image is smaller or bigger than the size
of the object.​

Answer 1

Answer:

• Hope the concept is clear!

Explanation:

$\underline{\pmb{\frak{Given:}}}$

• Object distance of a lens(u) = -30cm
• Image distance of a lens(v) = -10cm

Since the distance of the image is negative, the type of lens is nothing but a concave lens.

$\underline{\pmb{\frak{To\:find:}}}$

$\green{\bigstar}$ Magnification of the lens.

$\green{\bigstar}$ Whether the size of the image is smaller or bigger than the size of the object.

$\underline{\pmb{\sf{Formula\:for\:magnification:}}}$

$\maltese{\underline{\boxed{\pink{\sf{m = \dfrac{v}{u}}}}}}$

$\underline{\pmb{\frak{Solution:}}}$

On applying the magnification of lens formula, we get:

$\implies\bf{m = \dfrac{-10}{-30}}$

$\implies\bf{m = \dfrac{10}{30}}$

$\looparrowright{\pink{\tt{m = \dfrac{1}{3}}}}$

$\looparrowright{\pink{\tt{m = +0.3}}}$

Since, the magnification of the lens is positive, the image formed is virtual and erect.

The magnification of the lens is less than +1. This indicates that:

• Size of the Image is smaller than that of the object.

$\underline{\pmb{\frak{Remember:}}}$

• The image formed by a concave lens is always virtual, erect and diminished.