Power of Spherical Mirrors

What is the Power of a Mirror?

Imagine you have two concave mirrors. One brings parallel light rays to a focus just 10 cm away, while the other focuses them 50 cm away. Intuitively, the first mirror is "stronger" because it bends the light more sharply. This "strength" or "bending ability" is what we quantify as power.

Power (P) is defined as the measure of a mirror's ability to converge or diverge light rays.

• A mirror that causes a high degree of convergence or divergence has high power.
• A mirror that bends light rays only slightly has low power.

Since a mirror with a short focal length (f) bends light more strongly, power is inversely related to focal length.

The Formula and a Crucial Sign Convention

The formula for the power of a lens is P = 1/f. For a mirror, a specific convention is used to keep the signs consistent across all optical devices:

The power of a spherical mirror is given by:

Where:

• $P$ is the power in Dioptres (D).
• $f$ is the focal length, which must be in meters (m). This is a very common point of error, so always remember to convert f from cm to m!

Why the negative sign? This convention ensures that converging optical devices (like a concave mirror) always have positive power, and diverging devices (like a convex mirror) always have negative power. Let's see how this works.

Calculating Power for Concave and Convex Mirrors

Let's apply this formula using the sign conventions you've already mastered. To understand the relationship between mirror type, focal length, and power, let's turn to a concise resource.

After reading, let's consolidate the rules.

1. Concave Mirror (Converging Mirror)

• Sign of Focal Length: As you know, a concave mirror has a real focus. According to the Cartesian sign convention, its focal length ($f$) is negative.
• Calculating Power: $$P=-\frac{1}{f}=-\frac{1}{(\text{negative value})}$$
• Result: The power ($P$) of a concave mirror is positive.

Conclusion: A converging mirror has positive power.

2. Convex Mirror (Diverging Mirror)

• Sign of Focal Length: A convex mirror has a virtual focus behind the mirror. According to the Cartesian sign convention, its focal length ($f$) is positive.
• Calculating Power: $$P=-\frac{1}{f}=-\frac{1}{(\text{positive value})}$$
• Result: The power ($P$) of a convex mirror is negative.

Conclusion: A diverging mirror has negative power.

Here is a visual reminder of the sign convention for focal lengths that we use as the input for our power calculation.

Problem-Solving Practice

Let's work through a few NEET-style problems that integrate today's topic with what you learned in the previous lesson.

Problem 1: Direct Calculation

A concave mirror has a radius of curvature of 40 cm. Find its power.

Solution:

1. Find the focal length (f):
   $f=R/2=40\text{ cm}/2=20\text{ cm}$.
2. Apply the sign convention: It's a concave mirror, so the focal length is negative.
   $f=-20\text{ cm}$.
3. Convert to meters:
   $f=-0.2\text{ m}$.
4. Calculate Power (P): $$P=-\frac{1}{f}=-\frac{1}{-0.2\text{ m}}=+5.0\text{ D}$$ The power of the mirror is +5.0 Dioptres.

Problem 2: Working Backwards

The power of a spherical mirror is -2.5 D. Identify the type of mirror and find its focal length.

Solution:

1. Identify the mirror type: The power is negative, which corresponds to a diverging mirror. Therefore, it is a convex mirror.
2. Calculate Focal Length (f): $$P=-\frac{1}{f}⟹f=-\frac{1}{P}$$ $$f=-\frac{1}{-2.5\text{ D}}=+0.4\text{ m}$$
3. Convert to cm (optional but common):
   $f=+40\text{ cm}$.
   The focal length is +40 cm, and the positive sign confirms it is a convex mirror.

Problem 3: Multi-step Application

An object is placed in front of a spherical mirror, and a real image that is three times the size of the object is formed at a distance of 30 cm from the object. Calculate the power of the mirror.

Solution: This is a great problem because you have to deduce all the parameters first.

1. Analyze the given information:
   • "real image": Image is in front of the mirror.
   • "three times the size": Magnification magnitude is 3. Since the image is real, it must be inverted, so $m=-3$.
   • "formed at a distance of 30 cm from the object": This means the separation between $u$ and $v$ is 30 cm. Since a concave mirror is the only spherical mirror that forms a real, magnified image, the object must be between C and F, and the image is formed beyond C. The image is farther from the mirror than the object. Let the object distance be $|u|=x$. Then the image distance is $|v|=x+30$.
2. Apply Sign Convention:
   • Object distance $u=-x$
   • Image distance $v=-(x+30)$ (real image)
3. Use the Magnification Formula to find x: $$m=-\frac{v}{u}$$ $$-3=-\frac{-(x+30)}{-x}=-\frac{x+30}{x}$$ $$-3x=-(x+30)⟹-3x=-x-30⟹-2x=-30⟹x=15\text{ cm}$$
4. Determine u and v:
   • Object distance $u=-x=-15\text{ cm}$
   • Image distance $v=-(x+30)=-(15+30)=-45\text{ cm}$
5. Use the Mirror Formula to find f: $$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{1}{-45}+\frac{1}{-15}=\frac{-1-3}{45}=\frac{-4}{45}$$ $$f=-\frac{45}{4}=-11.25\text{ cm}$$
6. Convert f to meters:
   $f=-0.1125\text{ m}$
7. Calculate Power: $$P=-\frac{1}{f}=-\frac{1}{-0.1125}\approx +8.89\text{ D}$$ The power of the mirror is approximately +8.9 D.

Conclusion

Excellent work today! You have now learned the final theoretical concept for single spherical mirrors. By understanding power, you can quickly classify a mirror and quantify its fundamental optical property.

Key Takeaways:

• The power of a mirror ($P$) is a measure of its ability to converge or diverge light.
• The governing formula is $P=-1/f$, where $f$ must be in meters and $P$ is in Dioptres (D).
• The negative sign in the formula is a convention to ensure a consistent result:
  • Converging mirrors (concave) have a negative $f$ and thus positive power.
  • Diverging mirrors (convex) have a positive $f$ and thus negative power.

Preview of the Next Lesson:
You have now assembled a complete problem-solving toolkit for single mirrors: ray diagrams, sign conventions, the mirror formula, magnification, and power. In our next lesson, we will begin the "vigorous" problem-solving you're looking for. We will tackle a variety of complex, NEET-style questions that might involve finding unknown parameters, dealing with moving objects, or analyzing systems with more than one mirror. The solid foundation you've built will be essential.