A concave lens has a focal length of $15 ext{ cm}$. At what distance should the object from the lens be placed so that it forms an image at $10 ext{ cm}$ from the lens? Also, find the magnification produced by the lens. Show complete step-by-step calculations with proper sign conventions.

Q: A 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 complete step-by-step calculations with proper sign conventions.

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 ($m$) = ? Step 1: Finding the Object Distance ($u$)\nWe use the lens formula: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$| \nRearranging the formula to solve for object distance ($u$): $$\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$| \nSubstitute the given values with proper sign conventions: $$\frac{1}{u} = \frac{1}{-10} - \frac{1}{-15}$$ $$\frac{1}{u} = -\frac{1}{10} + \frac{1}{15}$| \nFind the least common multiple (LCM) of $10$ and $15$, which is $30$: $$\frac{1}{u} = \frac{-3 + 2}{30}$| $$\frac{1}{u} = \frac{-1}{30}$| $$u = -30\text{ cm}$| Conclusion for Object Distance: The object is placed at a distance of $30\text{ cm}$ in front of the concave lens. Step 2: Finding the Magnification ($m$)\nThe magnification for a lens is given by the formula: $$m = \frac{v}{u}$| \nSubstitute the values of $v$ and $u$: $$m = \frac{-10\text{ cm}}{-30\text{ cm}}$$ $$m = +\frac{1}{3} = +0.33$$ Conclusion for Magnification: The positive sign indicates that the image formed is virtual and erect. The magnitude of $0.33$ indicates that the image is diminished to one-third the size of the object.

Question

A concave lens has a focal length of $15	ext{ cm}$. At what distance should the object from the lens be placed so that it forms an image at $10	ext{ cm}$ from the lens? Also, find the magnification produced by the lens. Show complete step-by-step calculations with proper sign conventions.

Accepted Answer

Given Data: Focal length of the concave lens ($f$) = $-15	ext{ cm}$ (Focal length of a concave lens is always negative) Image distance ($v$) = $-10	ext{ cm}$ (A concave lens always forms a virtual, erect image on the same side as the object) Object distance ($u$) = ? Magnification ($m$) = ? Step 1: Finding the Object Distance ($u$)
We use the lens formula: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$| 
Rearranging the formula to solve for object distance ($u$): $$\frac{1}{u} = \frac{1}{v} - \frac{1}{f}$| 
Substitute the given values with proper sign conventions: $$\frac{1}{u} = \frac{1}{-10} - \frac{1}{-15}$$ $$\frac{1}{u} = -\frac{1}{10} + \frac{1}{15}$| 
Find the least common multiple (LCM) of $10$ and $15$, which is $30$: $$\frac{1}{u} = \frac{-3 + 2}{30}$| $$\frac{1}{u} = \frac{-1}{30}$| $$u = -30	ext{ cm}$| Conclusion for Object Distance: The object is placed at a distance of $30	ext{ cm}$ in front of the concave lens. Step 2: Finding the Magnification ($m$)
The magnification for a lens is given by the formula: $$m = \frac{v}{u}$| 
Substitute the values of $v$ and $u$: $$m = \frac{-10	ext{ cm}}{-30	ext{ cm}}$$ $$m = +\frac{1}{3} = +0.33$$ Conclusion for Magnification: The positive sign indicates that the image formed is virtual and erect. The magnitude of $0.33$ indicates that the image is diminished to one-third the size of the object.