Mastering Mirror Combinations: NEET Problem-Solving

Strategy 1: Tracing Rays in Mirror Combinations

The simplest combination involves two plane mirrors. The key here is a rigorous application of the law of reflection ($i=r$) and basic geometry.

Let's look at a typical NEET-style problem.

In this problem, a ray strikes Mirror 1. After reflecting, it strikes Mirror 2. We need to find the angle of reflection, $\theta$, from Mirror 2.

To solve this, we trace the path of the ray and analyze the angles at each step.

1. Reflection at Mirror 1: The angle of incidence is given as 55°. By the law of reflection, the angle of reflection is also 55°.
2. Geometry: Look at the triangle formed by the point of incidence on Mirror 1, the point of incidence on Mirror 2, and the vertex where the mirrors meet.
   • The angle between the reflected ray from Mirror 1 and Mirror 1 itself is $90^{\circ }-55^{\circ }=35^{\circ }$.
   • The angle between the mirrors is given as 120°.
   • The sum of angles in a triangle is 180°. So, the angle 'x' that the ray makes with Mirror 2 is: $x=180^{\circ }-120^{\circ }-35^{\circ }=25^{\circ }$.
3. Reflection at Mirror 2: The angle 'x' is the angle the ray makes with the mirror surface. The angle of incidence on Mirror 2 is the angle between the incoming ray and the normal to Mirror 2.
   • Angle of incidence at Mirror 2 = $90^{\circ }-x=90^{\circ }-25^{\circ }=65^{\circ }$.
4. Final Answer: By the law of reflection, the angle of reflection $\theta$ from Mirror 2 is equal to the angle of incidence, which is 65°.

This diagram shows the general geometry for such problems.

This diagram shows the various angles involved in successive reflections from two inclined mirrors. Mastering this geometry is key.

Shortcut: Number of Images

A common question asks for the total number of images formed by two inclined plane mirrors. While you can find this by tracing images, there's a much faster method for exams.

Summary of the Trick:

1. Calculate $n=\frac{360^{\circ }}{\theta }$, where $\theta$ is the angle between the mirrors.
2. If $n$ is an even integer: The number of images is $N=n-1$.
3. If $n$ is an odd integer: The number of images is $N=n$. (This applies when the object is placed symmetrically on the angle bisector, which is the standard case for most problems).

Strategy 2: Tackling Complex Scenarios with a Single Mirror

Even a single mirror can present complex challenges. Let's look at two common types.

A. Longitudinal Magnification: Object Along the Axis

What happens when an object isn't a small point perpendicular to the axis, but a rod lying along the principal axis?

The key is to treat the two ends of the object separately.

1. Find the image position ($v_A$) for the closer end (A).
2. Find the image position ($v_B$) for the farther end (B).
3. The length of the image is the distance between these two image points: $L_{image}=|v_A-v_B|$.

Let's solve the problem shown: A rod of length 10 cm lies along the principal axis of a concave mirror (f = -10 cm). Its closer end is 20 cm from the mirror.

• End A (closer end): $u_A=-20\text{ cm}$.
  Since this is at the center of curvature ($u=2f$), the image is also formed at C. So, $v_A=-20\text{ cm}$.

• End B (farther end): The rod is 10 cm long, so end B is at $20+10=30\text{ cm}$ from the mirror. $u_B=-30\text{ cm}$.
  Using the mirror formula:
  $\frac{1}{-10}=\frac{1}{v_B}+\frac{1}{-30}⟹\frac{1}{v_B}=\frac{1}{30}-\frac{1}{10}=\frac{1-3}{30}=\frac{-2}{30}=\frac{-1}{15}$
  So, $v_B=-15\text{ cm}$.

• Length of Image: The image of end A is at -20 cm, and the image of end B is at -15 cm.
  $L_{image}=|-20-(-15)|=|-5|=5\text{ cm}$.

Notice that the magnification is not uniform along the length of the rod. This is called longitudinal magnification.

B. Moving Objects and Mirrors: Image Velocity

Problems involving moving objects or mirrors can seem daunting, but they become straightforward with the right formula.

Key Formulas for Image Velocity:

1. Motion Perpendicular to the Mirror: The velocities of the image ($v_i$), object ($v_o$), and mirror ($v_m$) are related by:
   $2v_m=v_o+v_i$

   • Crucially, this is a vector equation. You must use sign conventions. A common choice is to take motion to the right as positive and to the left as negative.

2. Motion Parallel to the Mirror:

   • The image's parallel velocity component is equal to the object's parallel velocity component.
   • The mirror's parallel motion has no effect on the image's position or velocity.

Strategy 3: The Universal Principle for Combinations

When you have a system with multiple optical events (e.g., reflection then refraction, or reflection then another reflection), there is one guiding principle:

The image formed by one optical element serves as the object for the next element.

Let's see this in action.

Application 1: Mirror at the Bottom of a Liquid

This common scenario combines refraction (due to apparent depth) and reflection.

The Logic:

1. Direct View (Refraction only): An object at a real depth $d_{real}$ in a medium with refractive index $\mu$ appears to be at an apparent depth $d_{apparent}=d_{real}/\mu$.
2. Viewing the Reflection (Refraction -> Reflection -> Refraction):
   • Step A: A ray from the object travels to the mirror.
   • Step B: The plane mirror forms an image at the same distance behind it as the object is in front of it.
   • Step C: This image formed by the mirror now acts as a virtual object for the liquid surface.
   • Step D: An observer in the air sees this virtual object at an apparent depth calculated using the refraction formula. The total depth of this virtual object must be used as the "real depth" in the formula.

Application 2: Silvered Lenses

A silvered lens is a lens where one surface is coated to act as a mirror. This system acts as a single, effective mirror.

The Logic: A ray passes through the lens (refraction), reflects off the silvered surface (reflection), and passes back through the lens (refraction).
The effective power of this combination is the sum of the powers of each event:
$P_{eff}=P_{lens}+P_{mirror}+P_{lens}=2P_{lens}+P_{mirror}$

Since the final system is a mirror, its effective power is $P_{eff}=-1/f_{eff}$.
Remembering that $P_{lens}=1/f_{lens}$ and $P_{mirror}=-1/f_{mirror}$, we get:
$-\frac{1}{f_{eff}}=2\left(\frac{1}{f_{lens}}\right)+\left(-\frac{1}{f_{mirror}}\right)$
This formula allows you to find the effective focal length of the combination, which then behaves just like a single spherical mirror.

Conclusion

You have now equipped yourself with the strategies needed to deconstruct and solve the complex mirror problems you'll face in the NEET exam. The physics hasn't changed, but the application is more layered.

Key Takeaways:

• For combinations, the core principle is: the image from one element is the object for the next.
• For inclined plane mirrors, you can use geometry and the law of reflection or a shortcut formula for the number of images.
• For extended objects along the axis, calculate the image for each end point separately.
• For moving objects/mirrors, use the velocity formula $2v_m=v_o+v_i$, remembering to use sign conventions.
• For silvered lenses, treat the system as a single effective mirror whose power is $P_{eff}=2P_{lens}+P_{mirror}$.

Preview of the Next Lesson:
We are now ready to leave the world of reflection and move on to the next major topic in optics: Refraction. We will begin by exploring what happens when light passes from one medium to another, and we will derive and apply the fundamental law that governs this phenomenon: Snell's Law.