Question

11. Aconcave lens of focal length 15 cm forms an image 10 cm from the lens. How far
is the object placed from the lens? Draw the ray diagram.​

Answer 1

Answer :-

$\implies \boxed{ \sf{ u = - 30 \: cm \: }} \:$

→ Object is placed at 30 cm .

To Find :-

→ Object distance (u).

Explanation :-

Given that

• Focal length (f) = -15 cm ( left )

• image distance (v) = -10 (left)

• object distance (u) = ?

We know that ,

→ The lens formula due to a concave lens

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

substitute the given values .

$\implies \sf{\frac{1}{ - 15} = \frac{1}{ - 10} - \frac{1}{u} } \\ \\ \implies \sf{ \frac{1}{u} = \frac{1}{15} - \frac{1}{10} } \\ \\ \implies \sf{ \frac{1}{u} = \frac{2 - 3}{30} } \\ \\ \implies \sf{ \frac{1}{u} = \frac{ - 1}{30} } \\ \: \\ \implies \boxed{ \sf{ u = - 30 \: cm \: }}$

hence ,object is placed at a distance of 30 cm . negative sign shows that our object is on left side .

Answer 2

$\tt{\huge{\orange {Hello!!! }}}$

QUESTION :

A concave lens of focal length 15 cm forms an image 10 cm from the lens.

How far is the object placed from the lens?

Draw the ray diagram.

$\tt{\orange{---------}}$

SOLUTION :

$\tt{\red{\leadsto{Given \::- }}}$

The focal length is 15 cm.

The image distance is 10 cm.

$\tt{\blue{\mapsto{Formulae \: Used \::- }}}$

$\tt{\purple{\leadsto{ \dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v} }}}$

Now substituting the given Values into the above formula, :

$\tt{\red{\leadsto{ \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{-1}{10} + \dfrac{1}{15} = \dfrac{-1}{30} }}}$

$\tt{\pink{\implies{Object \: Distance \: = \: -30 \: cm. }}}$

Hence the distance of the object from the lens is -30 cm.

$\tt{\orange{---------}}$