Question

a convex lens has focal length 15cm.if the object placed at a distance of 12cm from the lens.where will the image formed and what kind of image is it?​

Answer 1

$\Large{\underline{\underline{\mathfrak{\red{\bf{Solution}}}}}}$

$\Large{\underline{\mathfrak{\orange{\bf{Given}}}}}$

• Focal length (f) = 15 cm
• Object Distance (u) = -12 cm

$\Large{\underline{\mathfrak{\orange{\bf{Find}}}}}$

• Distance of image (v)
• Nature of image

$\Large{\underline{\underline{\mathfrak{\red{\bf{Explanation}}}}}}$

Using Lens Formula

$\small{\boxed{\boxed{\tt{\red{\:\frac{1}{f}\:=\:\frac{1}{v}\:-\:\frac{1}{u}}}}}}$

Keep value of f & u

==> 1/15 = 1/v - 1/(-12)

==> 1/v = 1/15 - 1/12

==> 1/v = (12-15)/180

==> 1/v = -3/180

==> 1/v = -1/60

==> v = -60

$\Large{\underline{\underline{\mathfrak{\red{\bf{Hence}}}}}}$

• Distance of image be (v) = 60 cm
• Image formed in front of lens , it means image is virtual .

__________________

Answer 2

Solution :

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

• Focal length of lens, (f) = 15 cm
• Distance of object from lens, (u) = -12 cm

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

As we know that formula of the lens;

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

A/q

$\longrightarrow\sf{\dfrac{1}{15} =\dfrac{1}{v} -\dfrac{1}{(-12)}} \\\\\\\longrightarrow\sf{\dfrac{1}{15} =\dfrac{1}{v} +\dfrac{1}{12}}\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{15} -\dfrac{1}{12}}\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{4 - 5}{60}}\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{-1}{60}}\\\\\longrightarrow\sf{-v=60}\\\\\longrightarrow\bf{v=-60\:cm}$

Thus;

Image is formed at a distance of 60 cm in front of the mirror.

It's inverted & real .