Question

An optical flat is used to measure the height difference between a reference slip gauge A and a slip gauge B. Upon viewing via the optical flat using a monochromatic light of wavelength 0.5 \(\mu\)m, 12 fringes were observed over a length of 15 mm of gauge B. If the gauges are placed 45 mm apart, the height difference of the gauges is ............... \(\mu\)m. (Answer in integer) 

Hint:

The key relationship in optical flat measurements is that one fringe spacing corresponds to a height change of \(\lambda/2\). From this, you can find the slope of the air wedge and use it to determine any unknown height or length by simple proportion.

Answer 1

Step 1: Understanding the Concept:
Interference fringes are created when light waves reflect from the bottom surface of the optical flat and the top surface of the gauge block interfere. Each dark fringe corresponds to a location where the air gap thickness is an integer multiple of half the wavelength (\(n \lambda/2\)). The distance between two consecutive dark fringes corresponds to a change in height of \(\lambda/2\).
Step 2: Key Formula or Approach:
1. Calculate the height difference (\(h_{fringe}\)) corresponding to the total number of fringes observed over a certain length. \[ h_{fringe} = N \times \frac{\lambda}{2} \] where \(N\) is the number of fringes and \(\lambda\) is the wavelength of light. 2. The optical flat rests on the two gauges, forming a linear air wedge. We can use the principle of similar triangles to relate the height difference over the gauge width to the total height difference between the gauges. \[ \frac{\text{Total Height Difference } (\Delta H)}{\text{Total Distance between Gauges } (L_{total})} = \frac{\text{Height over Gauge B } (h_B)}{\text{Length on Gauge B } (L_B)} \] Step 3: Detailed Calculation:
1. Given Data:
- Wavelength, \(\lambda = 0.5 \, \mu\text{m}\).
- Number of fringes, \(N = 12\).
- Length over which fringes are observed, \(L_B = 15\) mm.
- Distance between gauges, \(L_{total} = 45\) mm.
2. Calculate the height change over gauge B:
The 12 fringes observed over the 15 mm length of gauge B correspond to a change in the air gap height.
\[ h_B = N \times \frac{\lambda}{2} = 12 \times \frac{0.5 \, \mu\text{m}}{2} = 12 \times 0.25 \, \mu\text{m} = 3 \, \mu\text{m} \] 3. Use similar triangles to find the total height difference (\(\Delta H\)):
Let \(\Delta H\) be the height difference between gauge A and gauge B.
\[ \frac{\Delta H}{L_{total}} = \frac{h_B}{L_B} \] \[ \frac{\Delta H}{45 \text{ mm}} = \frac{3 \, \mu\text{m}}{15 \text{ mm}} \] \[ \Delta H = \frac{3 \, \mu\text{m}}{15 \text{ mm}} \times 45 \text{ mm} \] \[ \Delta H = 3 \, \mu\text{m} \times \left(\frac{45}{15}\right) = 3 \, \mu\text{m} \times 3 = 9 \, \mu\text{m} \] Step 4: Final Answer:
The height difference of the gauges is 9 \(\mu\)m.
Step 5: Why This is Correct:
The solution correctly relates the number of interference fringes to the change in height of the air wedge. Then, using the linearity of the wedge (as the optical flat is flat), it correctly scales this height difference using similar triangles to find the total height difference over the full separation distance of the gauges.