0.5
Understanding Magnification in Concave Lenses
The question asks for the magnification produced by a concave lens given its focal length and the image distance. To find the magnification, we first need to determine the object distance using the lens formula, and then use the magnification formula.
Given Information and Sign Conventions
For a concave lens, the focal length is always negative. The image formed by a concave lens is always virtual, erect, and formed on the same side of the lens as the object. Therefore, the image distance is also taken as negative.
- Focal length of the concave lens, $f = -10$ cm (Negative for a concave lens)
- Image distance, $v = -5$ cm (Negative as the image is virtual and on the same side as the object)
We need to find the magnification, $m$.
Applying the Lens Formula
The lens formula relates the focal length ($f$), object distance ($u$), and image distance ($v$):
\(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\)
We can rearrange this formula to solve for the object distance, $u$:
\(\frac{1}{u} = \frac{1}{v} - \frac{1}{f}\)
Now, substitute the given values into the formula:
\(\frac{1}{u} = \frac{1}{-5 \text{ cm}} - \frac{1}{-10 \text{ cm}}\)
\(\frac{1}{u} = -\frac{1}{5 \text{ cm}} + \frac{1}{10 \text{ cm}}\)
To combine the fractions on the right side, find a common denominator, which is 10:
\(\frac{1}{u} = \frac{-2}{10 \text{ cm}} + \frac{1}{10 \text{ cm}}\)
\(\frac{1}{u} = \frac{-2 + 1}{10 \text{ cm}}\)
\(\frac{1}{u} = \frac{-1}{10 \text{ cm}}\)
Therefore, the object distance is:
\(u = -10\) cm
The negative sign for $u$ indicates that the object is placed on the left side of the lens, which is the standard convention for real objects.
Calculating Magnification
The magnification ($m$) produced by a lens is given by the ratio of the image distance ($v$) to the object distance ($u$):
\(m = \frac{v}{u}\)
Substitute the values of $v$ and the calculated $u$:
\(m = \frac{-5 \text{ cm}}{-10 \text{ cm}}\)
\(m = \frac{5}{10}\)
\(m = 0.5\)
The magnification is 0.5. A positive magnification indicates that the image is erect and virtual, which is consistent with the properties of a concave lens forming a virtual image. A magnification less than 1 (0.5 in this case) indicates that the image is diminished (smaller than the object).
Conclusion
The magnification produced by the concave lens is 0.5.
Revision Table: Concave Lens Formulas
| Quantity | Formula | Sign Conventions (Standard) |
|---|---|---|
| Lens Formula | \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\) | $f$: negative for concave lens $u$: negative for real object $v$: negative for virtual image (concave lens) |
| Magnification (m) | \(m = \frac{v}{u}\) | $m$: positive for erect/virtual image $m$: negative for inverted/real image $|m| < 1$: diminished image $|m| = 1$: same size image $|m| > 1$: magnified image |
Additional Information: Properties of Concave Lenses
A concave lens is a diverging lens. It is thinner at the center and thicker at the edges. Here are some key points about concave lenses:
- They always form virtual, erect, and diminished images for real objects.
- The image is always located between the optical center and the focal point on the same side as the object.
- They are used in eyeglasses to correct nearsightedness (myopia).
- Parallel rays of light incident on a concave lens diverge after refraction, appearing to originate from the principal focus on the same side.
- Rays passing through the optical center go undeviated.
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