# Article on Geometric and Physical Optics

## Introduction to Optics

Optics is the branch of physics that deals with the behavior and properties of light, including its interactions with matter and the instruments used to detect it. It encompasses both light’s visible spectrum and other forms of radiation such as ultraviolet and infrared light. Optics is divided into two main branches: geometric optics and physical optics.

## Geometric Optics

Geometric optics, or ray optics, involves the approximation of light as a collection of rays that travel in straight lines and bend when they encounter an interface between different materials. This model explains phenomena like reflection and refraction, which occur because light changes direction at interfaces. The laws of geometric optics can be understood using three fundamental principles:

1. **The Law of Reflection:** States that the angle of incidence is equal to the angle of reflection.

2. **The Law of Refraction (Snell’s Law):** Relates the angles of incidence and refraction to the indices of refraction of the two media.

3. **The Lens-Maker’s Equation:** Describes the relationship between the curvature of a lens’s surfaces and its focal length.

In geometric optics, light is often simplified to rays that can be analyzed using rules and principles without considering the wave nature of light.

## Physical Optics

Physical optics, on the other hand, deals with the wave properties of light. This branch focuses on the phenomena that cannot be explained by geometric optics, such as diffraction, interference, and polarization. In physical optics, light is understood as an electromagnetic wave, and wave equations are used to describe how light waves propagate.

Three key phenomena in physical optics include:

1. **Interference:** Occurs when two or more waves superimpose to form a resultant wave.

2. **Diffraction:** Describes how waves spread out as they pass through an opening or around obstacles.

3. **Polarization:** Refers to the orientation of the oscillations of the electric field vector in a light wave.

## Problems and Solutions in Geometric and Physical Optics

### Geometric Optics Problems

1. **Problem:** A 5 cm tall object is placed 30 cm from a concave mirror with a focal length of 15 cm. Calculate the image distance and the size of the image.

**Solution:**

Using the mirror equation:

\(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\)

Where \(f\) is the focal length, \(d_o\) is the object distance, and \(d_i\) is the image distance. Substituting the given values, we get:

\(\frac{1}{15} = \frac{1}{30} + \frac{1}{d_i}\)

Solving for \(d_i\):

\(\frac{1}{d_i} = \frac{1}{15} – \frac{1}{30} = \frac{1}{30}\)

Therefore, \(d_i = 30\) cm. Since the mirror is concave and the object is located beyond the focal point, the image is real and inverted. To find the image size (\(h_i\)), we use the magnification equation:

\(m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}\)

Substituting in the known values, we get:

\(-\frac{30}{30} = \frac{h_i}{5}\)

Solving for \(h_i\), we find:

\(h_i = -5\) cm. The negative sign indicates that the image is inverted.

2. **Problem:** A light ray strikes a flat glass slab (n=1.5) at an angle of 40 degrees to the normal. Find the angle of refraction inside the glass.

**Solution:**

Using Snell’s Law:

\(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\)

Where \(n_1 = 1\), \(n_2 = 1.5\), and \(\theta_1 = 40^\circ\). We want to find \(\theta_2\), the angle of refraction. After substituting the values, we get:

\(\sin(\theta_2) = \frac{\sin(40^\circ)}{1.5}\)

Solving for \(\theta_2\):

\(\theta_2 = \arcsin\left(\frac{\sin(40^\circ)}{1.5}\right)\)

After calculating, \(\theta_2 \approx 25.4^\circ\).

3. **Problem:** An underwater swimmer looks up at the surface of the water at a 30-degree angle. What is the angle of incidence of the light at the water-air surface assuming the index of refraction for water is 1.33?

**Solution:**

We want to find the angle of incidence (\(\theta_1\)) at the water-air interface with \(\theta_2 = 30^\circ\) and \(n_2 = 1\) (air). Since the light is coming from water to air, we use Snell’s Law in reverse:

\(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\)

Substituting values:

\(1.33 \sin(\theta_1) = \sin(30^\circ)\)

Solving for \(\theta_1\):

\(\sin(\theta_1) = \frac{\sin(30^\circ)}{1.33}\)

\(\theta_1 = \arcsin\left(\frac{1}{2 \cdot 1.33}\right)\)

Calculating, we find that \(\theta_1 \approx 21.8^\circ\).

4. **Problem:** A plano-convex lens has a radius of curvature of 20 cm. If the lens is made of material with a refractive index of 1.6, what is its focal length?

**Solution:**

For a plano-convex lens, the lens-maker’s equation is:

\(\frac{1}{f} = (n-1) \left( \frac{1}{R_1} – \frac{1}{R_2} \right)\)

Since one side is flat, \(R_2 = \infty\), and the equation simplifies to:

\(\frac{1}{f} = (n-1)\frac{1}{R_1}\)

Substituting the given values:

\(\frac{1}{f} = (1.6 – 1)\frac{1}{20}\)

Solving for \(f\), we get:

\(\frac{1}{f} = 0.6 \cdot \frac{1}{20}\)

\(f = \frac{20}{0.6}\)

Focal length, \(f \approx 33.3\) cm.

5. **Problem:** A bi-convex lens has a focal length of 50 cm. When an object is placed 75 cm from the lens, where is the image formed?

**Solution:**

Use the lens equation:

\(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\)

Substituting known values:

\(\frac{1}{50} = \frac{1}{75} + \frac{1}{d_i}\)

Solving for \(d_i\):

\(\frac{1}{d_i} = \frac{1}{50} – \frac{1}{75} = \frac{3}{150} – \frac{2}{150}\)

\(\frac{1}{d_i} = \frac{1}{150}\)

Therefore, \(d_i = 150\) cm.

### Physical Optics Problems

6. **Problem:** Two coherent light sources with a wavelength of 600 nm interfere creating a pattern with a fringe spacing of 0.5 mm on a screen 2 m away. What is the distance between the light sources?

**Solution:**

Fringe spacing (\(y\)) in an interference pattern is given by:

\(y = \frac{\lambda L}{d}\)

Where \(\lambda\) is the wavelength, \(L\) is the distance to the screen, and \(d\) is the distance between the light sources. Rearranging this formula to solve for \(d\):

\(d = \frac{\lambda L}{y}\)

Substituting the given values:

\(d = \frac{600 \times 10^{-9} \times 2}{0.5 \times 10^{-3}}\)

After calculation, we find:

\(d = 2.4 \times 10^{-3}\) m or 2.4 mm.

7. **Problem:** A single slit diffraction experiment creates a first minimum at an angle of 30 degrees. The wavelength of the light used is 500 nm. Calculate the width of the slit.

**Solution:**

The angle for the first minimum in a single slit diffraction pattern, \(\theta\), is given by:

\(a \sin(\theta) = m\lambda\)

For the first minimum, \(m = 1\). Rearranging to solve for the width of the slit (\(a\)):

\(a = \frac{\lambda}{\sin(\theta)}\)

Plugging in the values:

\(a = \frac{500 \times 10^{-9}}{\sin(30^\circ)}\)

Calculating the width, \(a \approx 1 \times 10^{-6}\) m or 1 µm.

8. **Problem:** What is the angular width of the central maximum in a single-slit diffraction pattern when a slit of width 0.02 mm is illuminated with a 400 nm light?

**Solution:**

The angular width of the central maximum is given by:

\(\Delta\theta = \frac{2\lambda}{a}\)

Substituting the given values:

\(\Delta\theta = \frac{2 \times 400 \times 10^{-9}}{0.02 \times 10^{-3}}\)

After calculation:

\(\Delta\theta = 0.04\) radians or approximately \(2.29^\circ\).

9. **Problem:** A polarizer reduces the intensity of polarized light to 50% of its original intensity. What is the angle between the light’s polarization direction and the polarizer’s transmission axis?

**Solution:**

When light passes through a polarizer, the intensity \(I\) after the polarizer is related to the initial intensity \(I_0\) and the angle \(\theta\) between the light’s polarization direction and the polarizer’s axis by Malus’s law:

\(I = I_0 \cos^2(\theta)\)

Given that \(I = 0.5I_0\), we can solve for \(\theta\):

\(0.5I_0 = I_0 \cos^2(\theta)\)

\(0.5 = \cos^2(\theta)\)

Which, when solved, gives:

\(\theta = \cos^{-1}(\sqrt{0.5})\)

This yields:

\(\theta = 45^\circ\) or \(\pi/4\) radians.

10. **Problem:** If a beam of unpolarized light with intensity \(I_0\) passes through three polarizers, with the second and third polarizers oriented at 60 degrees to the first and each other, respectively, what is the final intensity of the light?

**Solution:**

For unpolarized light passing through a polarizer:

\(I = \frac{I_0}{2}\)

After passing through the second polarizer at 60 degrees to the first:

\(I = \frac{I_0}{2} \cos^2(60^\circ)\)

The intensity after the third polarizer, oriented at 60 degrees to the second:

\(I = \frac{I_0}{2} \cos^2(60^\circ) \cos^2(60^\circ)\)

Substituting the cosine values:

\(I = \frac{I_0}{2} \left(\frac{1}{2}\right)^2\)

Simplifying:

\(I = \frac{I_0}{8}\)

Therefore, the final intensity is \(I_0/8\).

These 10 problems provide a mix of challenges in the areas of geometric and physical optics, showcasing the use of essential equations and principles from both fields. Remember, when resolving optics problems, it is crucial to carefully consider the nature of light, whether it is being treated as a ray in geometric optics or as a wave in physical optics.