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Plot a graph of y against sin i Find K = S' State two precautions taken to ensure accurate results. Distinguish between reflection and refraction of light. A ray of light is incident at an angle of 30° to the normal, is deviated through an angle of 53° in a rectangular prism.

Physics

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1. Plot a graph of y against sin i
To plot the graph, use the following data points where sin⁡i\sin i is on the x-axis and yy is on the y-axis:

  • Point 1: (sin⁡i,y)=(0.866,0.60)(\sin i, y) = (0.866, 0.60)
  • Point 2: (sin⁡i,y)=(0.766,0.53)(\sin i, y) = (0.766, 0.53)
  • Point 3: (sin⁡i,y)=(0.643,0.50)(\sin i, y) = (0.643, 0.50)
  • Point 4: (sin⁡i,y)=(0.500,0.40)(\sin i, y) = (0.500, 0.40)
  • Point 5: (sin⁡i,y)=(0.342,0.30)(\sin i, y) = (0.342, 0.30)

2. Find K=S
Assuming 'S' refers to the slope of the graph of yy against sin⁡i\sin i, we can calculate K. We will use the first two data points as they show a consistent relationship.

  1. Identify the coordinates of two points.

    Let (sin⁡i1,y1)=(0.866,0.60)(\sin i_1, y_1) = (0.866, 0.60) and (sin⁡i2,y2)=(0.766,0.53)(\sin i_2, y_2) = (0.766, 0.53).

  2. Calculate the slope K.

    K=y2−y1sin⁡i2−sin⁡i1K = \frac{y_2 - y_1}{\sin i_2 - \sin i_1} K=0.53−0.600.766−0.866K = \frac{0.53 - 0.60}{0.766 - 0.866} K=−0.07−0.100K = \frac{-0.07}{-0.100} K=0.70K = 0.70

    The value of K (the slope) is 0.70\boxed{\text{0.70}}.

    3. State two precautions taken to ensure accurate results
    Two precautions to ensure accurate results in an optics experiment are:

    • Ensure the optical components (e.g., prism, lens, pins) are clean and free from dust or smudges.
    • Use a sharp pencil to mark points on the paper to minimize errors in measurement and avoid parallax error when reading scales or angles.

    4. Distinguish between reflection and refraction of light

    • Reflection is the phenomenon where light bounces off a surface and returns into the same medium. The angle of incidence equals the angle of reflection.
    • Refraction is the phenomenon where light passes from one medium to another, causing it to change direction due to a change in speed. This bending of light is governed by Snell's Law.

    5. A ray of light incident at an angle of 30° to the normal is deviated through an angle of 10.53° in a rectangular prism.
    We will calculate the refractive index of the prism.

  3. Identify the given values.

    Angle of incidence, i=30∘i = 30^\circ.
    Angle of deviation, D=10.53∘D = 10.53^\circ.

  4. Calculate the angle of refraction rr.

    When light enters a denser medium (like a prism from air), it bends towards the normal. The deviation is given by the difference between the angle of incidence and the angle of refraction:

    D=i−rD = i - r 10.53∘=30∘−r10.53^\circ = 30^\circ - r r=30∘−10.53∘r = 30^\circ - 10.53^\circ r=19.47∘r = 19.47^\circ
  5. Apply Snell's Law to find the refractive index nn.

    Snell's Law states n1sin⁡i=n2sin⁡rn_1 \sin i = n_2 \sin r. Assuming light travels from air (n1=1n_1 = 1) into the prism (n2=nn_2 = n):

    1⋅sin⁡30∘=nsin⁡19.47∘1 \cdot \sin 30^\circ = n \sin 19.47^\circ n=sin⁡30∘sin⁡19.47∘n = \frac{\sin 30^\circ}{\sin 19.47^\circ} n=0.50.33333n = \frac{0.5}{0.33333} n≈1.500n \approx 1.500

    The refractive index of the prism is approximately 1.500\boxed{\text{1.500}}.

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