Mirrors, Refraction, and Reflection: Numerical Problems

1. Problem Solving: Spherical Mirrors

The most common source of error in mirror problems is the sign convention. Let's start with a quick and clear review of the rules.

To summarize the key conventions (using the Cartesian system):

• Light is assumed to travel from left to right.
• The pole (P) is the origin (0,0).
• Distances measured in the direction of incident light are positive; those measured against it are negative.
• Heights measured upwards from the principal axis are positive; downwards are negative.

This leads to:

• u (object distance) is almost always negative.
• f (focal length) is negative for concave mirrors and positive for convex mirrors.
• v (image distance) is negative for real images (in front of the mirror) and positive for virtual images (behind the mirror).
• m (magnification) is negative for real, inverted images and positive for virtual, erect images.

Example 1: Concave Mirror and Magnification

Let's solve a problem where we need to work backward from the magnification to find the object position.

Problem: Calculate the distance of an object from a concave mirror with a focal length of 10 cm, so that a real image is formed with a magnification of 2.

Let's outline the solution:

1. Given:

   • Mirror type: Concave $⟹f=-10$ cm.
   • Image type: Real $⟹$ magnification $m$ is negative.
   • Magnification magnitude is 2. So, $m=-2$.

2. Relate $v$ and $u$:

   • Use the magnification formula: $m=-\frac{v}{u}$.
   • $-2=-\frac{v}{u}⟹v=2u$.

3. Apply the Mirror Formula:

   • $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$
   • $\frac{1}{-10}=\frac{1}{2u}+\frac{1}{u}$
   • $\frac{1}{-10}=\frac{1+2}{2u}=\frac{3}{2u}$

4. Solve for $u$:

   • $2u=-30$
   • $u=-15$ cm.

The object must be placed 15 cm in front of the mirror.

2. Problem Solving: Refraction

Problems involving refraction typically use Snell's Law or the concept of apparent depth.

Example 2: Calculating Refractive Index

Problem: A ray of light travels a distance of 12 m in a transparent sheet in $60$ nanoseconds. The speed of light in vacuum is $3\times 10^8$ m/s. What is the refractive index of the sheet?

Here is the breakdown:

1. Given:

   • Distance in medium, $d=12$ m.
   • Time in medium, $t=60\text{ ns}=60\times 10^{-9}$ s.
   • Speed of light in vacuum, $c=3\times 10^8$ m/s.

2. Calculate Speed in Medium ($v$):

   • $v=\frac{d}{t}=\frac{12}{60\times 10^{-9}}=\frac{1}{5\times 10^{-9}}=0.2\times 10^9=2\times 10^8$ m/s.

3. Calculate Refractive Index ($n$):

   • $n=\frac{c}{v}=\frac{3\times 10^8}{2\times 10^8}=1.5$.

The refractive index of the sheet is 1.5.

3. Problem Solving: Total Internal Reflection (TIR)

These problems test your understanding of the two conditions for TIR and the critical angle formula.

Example 3: Critical Angle and Speed of Light

Problem: The critical angle for light going from a certain medium into air is $30^{\circ }$. Find the speed of light in the medium.

This problem connects the concept of critical angle back to the fundamental definition of refractive index.

Let's walk through the steps:

1. Given:

   • Light goes from medium to air, so $n_1=n$, $n_2=n_{air}\approx 1$.
   • Critical angle, $i_c=30^{\circ }$.

2. Find the Refractive Index ($n$):

   • Using the critical angle formula: $\sin i_c=\frac{n_2}{n_1}=\frac{1}{n}$.
   • $n=\frac{1}{\sin i_c}=\frac{1}{\sin 30^{\circ }}=\frac{1}{1/2}=2$.

3. Find the Speed of Light in the Medium ($v$):

   • Using the definition of refractive index: $n=\frac{c}{v}$.
   • $v=\frac{c}{n}=\frac{3\times 10^8\text{ m/s}}{2}=1.5\times 10^8$ m/s.

Example 4: A Challenging TIR Application (Snell's Window)

This final problem is a classic that beautifully integrates geometry with the physics of TIR.

Problem: A point source of light is placed at the bottom of a tank filled with water (refractive index $\mu$) to a depth of $d$. Show that the area of the water surface through which light can emerge into the air is a circle of area $\pi d^2/({\mu }^2-1)$.

Let's derive it together:

1. Conceptual Setup: Light radiates from the source in all directions. Light rays striking the surface at an angle $i<i_c$ will refract out. The ray striking at exactly $i=i_c$ will graze the surface. Any ray striking at $i>i_c$ will be totally internally reflected. Therefore, the emerging light is confined to a circle on the surface, the radius of which is defined by the critical angle.

2. Geometric Relation:

   • Let the radius of this circle be $R$. Let the depth of the source be $d$.
   • Consider the right-angled triangle formed by the light source, the center of the circle on the surface, and the edge of the circle. The angle at the source is the critical angle, $i_c$.
   • From trigonometry, $\tan i_c=\frac{\text{opposite}}{\text{adjacent}}=\frac{R}{d}$.
   • So, the radius is $R=d\tan i_c$.

3. Using the TIR Formula:

   • We know $\sin i_c=\frac{n_{air}}{n_{water}}=\frac{1}{\mu }$.
   • We need $\tan i_c$. We can use the identity $\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }}$.
   • $\tan i_c=\frac{1/\mu }{\sqrt{1-(1/\mu {)}^2}}=\frac{1/\mu }{\sqrt{({\mu }^2-1)/{\mu }^2}}=\frac{1/\mu }{\frac{\sqrt{{\mu }^2-1}}{\mu }}=\frac{1}{\sqrt{{\mu }^2-1}}$.

4. Find the Radius and Area:

   • Substitute this expression for $\tan i_c$ back into our equation for $R$:
     $R=d\tan i_c=\frac{d}{\sqrt{{\mu }^2-1}}$
   • The area of the circle is $A=\pi R^2$.
     $A=\pi {\left(\frac{d}{\sqrt{{\mu }^2-1}}\right)}^2=\frac{\pi d^2}{{\mu }^2-1}$
   • This completes the proof.

Conclusion

Today, we bridged the gap between theory and application by tackling a variety of numerical problems. You've seen how to systematically apply the core formulas of this module to find practical, quantitative answers.

• Key Takeaways:
  • A structured approach (diagram, list variables, check signs) is essential for accuracy.
  • Sign conventions are not arbitrary rules but a necessary framework for using the mirror and lens formulas correctly.
  • The principles of refraction and TIR can be combined with geometry to solve complex and interesting problems, like Snell's Window.

Preview of the Next Lesson:
With this module on reflection and refraction at plane and spherical surfaces complete, we are ready to study a key optical component: the prism. In our next lesson, we will begin Module 2 by analyzing the path of light through a prism, deriving the famous prism formula, and seeing how TIR is applied within prisms. The concepts of refraction we've practiced today are the direct foundation for that analysis.