Question

A concave lens of focal length 15 cm forms an image at a distance of 10 cm from the lens. How far is the object from the lens? What is its nature and magnification?​

Answer 1

S O L U T I O N :

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

• Focal length of the lens, (f) = -15 cm
• Image distance of the lens, (v) = -10 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

$\longrightarrow\tt{\dfrac{1}{f} =\dfrac{1}{v} -\dfrac{1}{u} }$

$\longrightarrow\tt{\dfrac{1}{-15} =\dfrac{1}{-10} -\dfrac{1}{u} }$

$\longrightarrow\tt{\dfrac{1}{-15} + \dfrac{1}{10} = -\dfrac{1}{u} }$

$\longrightarrow\tt{\dfrac{-2 + 3}{30} = -\dfrac{1}{u} }$

$\longrightarrow\tt{\dfrac{1}{30} = -\dfrac{1}{u} }$

$\longrightarrow\bf{u = -30\:cm}$

∴ The object distance will be 30 cm .

As we know that formula of the magnification;

$\boxed{\bf{Magnification,(m) = \frac{h'}{h} = \frac{v}{u} }}$

$\longrightarrow\tt{m = \dfrac{v}{u}}$

$\longrightarrow\tt{m = \dfrac{-10}{-30}}$

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

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

Thus,

The magnification will be 0.33 cm & the image is erect and virtual .

Answer 2

➱Focal length (f ) = -15 cm

➱Distance of image (v) = -10 cm

➱1/v–1/u=1/f

➱1/u = -(1/10) – (1/- 15)

➱1/u =1/15 – 1/10

➱1/u = -0.033

➱u = -30 cm

☞so the object is placed 30 cm away from the concave lens.

Ray diagram:-

⚠️refer to teh attachment⚠️☝️