Given tanθ=x/L,sinθ=mλ/d,λ=440 nm,m=4,L=0.6 m,x=5.3 cm find d in mm :

To find \( d \) using the given equations, we can start by understanding the relationships and solving the equations step-by-step:

### Given Data:

1. \( \tan\theta = \frac{x}{L} \)
2. \( \sin\theta = \frac{m\lambda}{d} \)
3. \( \lambda = 440 \) nm \( = 440 \times 10^{-9} \) m
4. \( m = 4 \)
5. \( L = 0.6 \) m
6. \( x = 5.3 \) cm \( = 0.053 \) m

### Steps to Solve for \( d \):

1. **Find \(\theta\) using \( \tan\theta \):**

   \[
   \tan\theta = \frac{x}{L} = \frac{0.053 \text{ m}}{0.6 \text{ m}}
   \]

   \[
   \tan\theta = \frac{0.053}{0.6} \approx 0.08833
   \]

   To find \(\theta\), we take the arctan (inverse tangent):

   \[
   \theta = \arctan(0.08833)
   \]

2. **Calculate \(\theta\):**

   \[
   \theta \approx 5.05^\circ \text{ (in degrees)}
   \]
   
   Converting \(\theta\) into radians for further calculations:

   \[
   \theta \approx \arctan(0.08833) \approx 0.0882 \text{ radians}
   \]

3. **Use the value of \(\theta\) to find \( d \) from \( \sin\theta = \frac{m\lambda}{d} \):**

   \[
   \sin\theta \approx \sin(0.0882) \approx 0.0882
   \]

   Substituting into the second equation:

   \[
   0.0882 = \frac{m\lambda}{d}
   \]

4. **Rearrange to solve for \( d \):**

   \[
   d = \frac{m\lambda}{\sin\theta}
   \]

   Substitute the known values:

   \[
   d = \frac{4 \times 440 \times 10^{-9} \text{ m}}{0.0882}
   \]

5. **Calculate \( d \):**

   \[
   d = \frac{1760 \times 10^{-9} \text{ m}}{0.0882}
   \]

   \[
   d \approx 1.994 \times 10^{-5} \text{ m}
   \]

   Converting \( d \) to millimeters:

   \[
   d \approx 0.01994 \text{ m} = 19.94 \text{ mm}
   \]

### Final Answer:

\[
d \approx 19.94 \text{ mm}
\]

Would you like more details or have any questions?

### Relative Questions

1. How would the value of \(d\) change if \( \lambda \) were increased to 500 nm?
2. What would happen to \(d\) if \(m\) were decreased to 3?
3. How does the distance \(L\) affect the value of \(\theta\) in these equations?
4. Can the relationship \( \tan\theta = \frac{x}{L} \) be used for very large values of \(\theta\)?
5. How does diffraction affect the interference pattern observed in this experiment?

### Tip:

When dealing with small angles in physics problems, the approximation \( \sin\theta \approx \tan\theta \approx \theta \) (in radians) can often simplify calculations.

Explore how to find the optical path difference using trigonometric and wave optics principles. Solve for the distance \( d \) given the wavelength, angle, and distances involved. Suitable for advanced high school students and physics enthusiasts.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation