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. Show all calculations clearly.

Numerical Problem Solution

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, erect image on the same side as the object)
• Object distance ($u$) = ?
• Magnification ($m$) = ?

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

Step-by-Step Calculation for Object Distance ($u$):

1. Rearrange the lens formula to solve for $\frac{1}{u}$: $$\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$|
2. Substitute the given values with proper sign conventions into the equation: $$\frac{1}{u} = \frac{1}{-10} - \frac{1}{-15}$| $$\frac{1}{u} = -\frac{1}{10} + \frac{1}{15}$|
3. Take the Least Common Multiple (LCM) of $10$ and $15$, which is $30$: $$\frac{1}{u} = \frac{-3 + 2}{30}$| $$\frac{1}{u} = \frac{-1}{30}$|
4. Invert both sides to find $u$: $$u = -30 \text{ cm}$|

Formula 2: Magnification for Lens $$m = \frac{v}{u}$|

Step-by-Step Calculation for Magnification ($m$):

1. Substitute the values of $v$ and $u$ into the magnification formula: $$m = \frac{-10 \text{ cm}}{-30 \text{ cm}}$|
2. Simplify the fraction: $$m = \frac{1}{3} \approx +0.33$$

Answer:

1. The object must be placed at a distance of $30 \text{ cm}$ in front of the concave lens (indicated by the negative sign).
2. The magnification produced by the lens is $+0.33$, indicating that the image is virtual, erect, and one-third the size of the object.