CBSE · Class 10 · Science · Light – Reflection and RefractionA concave lens has a focal length of $15\text{ cm}$. At what distance should the object from the lens be placed so that it forms an image at $10\text{ cm}$ from the lens? Also, find the magnification produced by the lens.

Given Data:

• Focal length of the concave lens ($f$) = $-15\text{ cm}$ (Focal length of a concave lens is always negative)
• Image distance ($v$) = $-10\text{ cm}$ (A concave lens always forms a virtual image on the same side as the object)

To find:

• Object distance ($u$) = ?
• Magnification ($m$) = ?

Formula:\nUsing the Lens Formula:

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$|

Step-by-Step Calculation:

1. Rearranging the lens formula to find $u$: $$\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$$
2. Substituting the given values: $$\frac{1}{u} = \frac{1}{-10} - \frac{1}{-15}$$ $$\frac{1}{u} = -\frac{1}{10} + \frac{1}{15}$$
3. Taking the LCM of 10 and 15, which is 30: $$\frac{1}{u} = \frac{-3 + 2}{30}$| $$\frac{1}{u} = \frac{-1}{30}$$ $$u = -30\text{ cm}$| \nSo, the object is placed at a distance of $30\text{ cm}$ in front of the concave lens.

Calculation of Magnification ($m$):\nThe formula for magnification for a lens is:

$$m = \frac{v}{u}$$\nSubstituting the values of $v$ and $u$: $$m = \frac{-10}{-30}$$ $$m = +\frac{1}{3} = +0.33$$

Conclusion:\nThe object is placed at a distance of $30\text{ cm}$ from the lens. The positive sign of magnification indicates that the image formed is virtual and erect, and its size is one-third the size of the object.