To find the image distance ($v$) and magnification ($m$) for a thin concave (diverging) lens, we can use the lens formula and the magnification formula.

1. Lens formula: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ where:
   • $f = -10$ cm (focal length for a concave lens is negative),
   • $u = -15$ cm (object distance is negative for real objects).

Plugging in the values: $\frac{1}{-10} = \frac{1}{v} + \frac{1}{-15}$ This simplifies to: $\frac{1}{v} = -\frac{1}{10} + \frac{1}{15}$ Finding a common denominator (30): $\frac{1}{v} = -\frac{3}{30} + \frac{2}{30} = -\frac{1}{30}$ Thus, $v = -30 \text{ cm}$

2. Magnification formula: $m = -\frac{v}{u}$ Substituting the values: $m = -\frac{-30}{-15} = 2$

Summary:

• Image distance $v = -30$ cm (indicating a virtual image on the same side as the object).
• Magnification $m = 2$ (the image is upright and twice the size of the object).

Do you want details or have any questions? Here are some related questions:

1. What is the significance of the sign convention in optics?
2. How does the magnification relate to the type of lens used?
3. Can you explain how to draw a ray diagram for this scenario?
4. What happens to the image characteristics if the object distance changes?
5. How does the focal length affect the image distance and magnification?

Tip: Always remember to apply the correct sign conventions for lens formulas to avoid confusion in calculations.