Physics Practical — Refractive Index of a Glass Block

Objectives: Physics Practical — Refractive Index of a Glass Block

Physics Practical — Refractive Index of a Glass Block (Q&A)

Physics Practical — Refractive index of a glass block (Q & A)

Question (practical)

Determine the refractive index n of a glass block using the pin-and-ray method. For several angles of incidence i (e.g. 25°, 35°, 45°, 55°, 65°) you measure the distance L (as described in the experiment). From the measurements you are asked to:

  1. Tabulate values of sin² i and 1 / L².
  2. Plot a graph of sin² i (vertical axis) against 1 / L² (horizontal axis).
  3. Find the slope S and intercept C of the straight line.
  4. Determine the refractive index using n = √C.
  5. Find the width w of the block from the relation w = √(-S) / n.

Short answer (result)

Using the sample data below we found:

  • Slope: S = -0.5000
  • Intercept: C = 2.2500
  • Refractive index: n = √C = 1.5000
  • Width of block: w = √(-S) / n = 0.4714 (same units as L)

Detailed solution (how to get S and C)

If the line equation is y = S·x + C where y = sin² i and x = 1/L², the slope S and intercept C can be found by linear least squares:

S = [ NΣ(xy) − Σx Σy ] / [ NΣ(x²) − (Σx)² ]
C = [ Σy − S Σx ] / N

(N is number of points). After computing S and C we get n = √C and w = √(-S) / n.

Worked numeric demonstration

Using the example data (angles i = 25°, 35°, 45°, 55°, 65°) and measured L values (in centimetres), we tabulated sin² i and 1 / L² and performed a straight-line fit. The computed best-fit slope and intercept are shown in the result above.

Notes for the experiment

  • Make sure pins P1 and P2 produce a clear apparent line seen through the face of the block.
  • Take each L measurement carefully and repeat each reading at least twice; average them.
  • Use the central axis of the block and avoid parallax when aligning pins.

SVG diagram (ray through a rectangular block)

A (front) B (front) C (back) D (back) P1 (pin) P4 O L i (angle of incidence) refracted ray

Figure: A front view of a glass block. Pins are placed so emergent ray is traced and distance L measured.

Sample data & computed table

Example data used for the worked solution (units of L are centimetres):

Angle i (°) sin² i Measured L (cm) 1 / L² (cm⁻²)
Note: values shown to 6 significant digits for clarity.

Linear fit (y = S·x + C)

From least squares fit we obtain: S = -0.5000 and C = 2.2500.

Compute refractive index
n = √C → n = √2.25 = 1.5000
Compute block width
w = √(-S) / n → √(-(-0.5)) / 1.5 = √0.5 / 1.5 = 0.70710678 / 1.5 = 0.47140452
Rounded: 0.4714 (same units as L)

Graph (SVG) — sin²i vs 1/L²

Blue dots are measured points; red line is the best-fit straight line with slope S and intercept C.

Interpretation: The intercept C equals n² (so n = √C). The slope is negative; its magnitude relates to the block width as shown above.

Complete step-by-step answer (for exam use)

  1. Draw the outline of the glass block and fix pins P₁ and P₂ such that the incident ray at the front face makes the required angle i with the normal.
  2. Place pins P₃ and P₄ so that they appear in a straight line with the images of P₁ and P₂ when viewed through the opposite face — measure the distance L between the point where the emergent ray crosses a reference line and the point O inside the block (use the experimental definition provided).
  3. Repeat for the requested angles i = 25°, 35°, 45°, 55°, 65°.
  4. For each i compute sin²i and 1/L² and tabulate the results.
  5. Plot sin²i (vertical) against 1/L² (horizontal), draw the best-fit straight line and determine slope S and intercept C.
  6. Compute n = √C and w = √(-S) / n. Report your final values with proper units and uncertainties (if you computed them).

Exam tip: show the formula linking n to C clearly and explain sign of S (S is usually negative for this arrangement); always give the unit for w (same as L, e.g. cm).

Reference Book: N/A

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