Anonymous
Anonymous asked in Science & MathematicsPhysics · 4 weeks ago

Can someone explain please?

You set up an experiment with two polarizer screens on an optical bench and collect data for the intensity of the light at the detector as you vary the relative angle between the transmission axes of the polarizers. You create a plot of light intensity (in lux) vs cos2 θ with your data and get a linear fit for your graph gives a slope of 120.2 ± 1.204 and a y-intercept of 15.45 ± 2.141.

(a) What is intensity of the light source you used in your experiment? (in lux)

(b) What is ambient light in the room as you took your data? (in lux)

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  • Vaman
    Lv 7
    4 weeks ago

    First polariser sends light linearly  polarised along the axis say it is along the y. Electric field E is along this axis. The intensity is propprtional to electric field^2. At any angle to tis axis, the electric field will be e= E cos a,, The intensity = e^2 = E^2cos^2 a= 1/2 E^2 (1+cos 2a). Intensity at an angle a will be I0 (1+cos 2a). I do not know the method to go ahead. So essentially you have the graph   I =I0 +I0 cos 2a. At a=0, I=2 I0= 15.45. Now you will have I=15.45/2(1+cos 2a). According to this calculation the intensity is 15.45 Lux. Intensity in the room is 5.3 using the formula 15.45*e^(-1.07).  I do not know whether this is the correct answer

  • 4 weeks ago

    The “slope of 120.2 ± 1.204“has mismatched precision and missing units. (It’s inappropriate to match significant figures in this situation.)  It could correctly be written as:

    (120.2 ± 1.2) lux, or

    (120.200 ± 1.204) lux

    Similarly the y-intercept could be written as either:

    (15.45 ± 2.14) lux, or

    (15.450 ± 2.141) lux

    So you need to correct these if the values are your own.   But see my note at the very end.

    __________________

    Incident intensity = I₀.

    Transmitted intensity = Iₜ

    Measured intensity = Iₘ

    Ambient light level = Iₐ

    Iₜ = I₀cos²θ  (law of Malus) and Iₘ = Iₜ.+  Iₐ (law of common sense). Combining these gives:

    Iₘ =I₀cos²θ + Iₐ (equation 1)

    Note, y = mx+c is a straight line, gradient m and y-intercept c.

    Comparing 'y = mx+c' to equation 1 you see they are basically the same:

    The variable y corresponds to the variable Iₘ.

    The constant m corresponds to the constant I₀.

    The variable corresponds to the variable cos²θ.

    The constant c corresponds to the constant Iₐ.

    So no calculation is needed.  You are actually given the answers!

    a) the intensity of the light source is I₀ = (120.2 ± 1.204) lux

    b) the ambient light level is Iₐ = (15.45 ± 2.141) lux

    I’d be surprised if the actual measurements are so accurate.  It depends what equipment you used.  We usually round uncertainty to 1 significant figure and then round the value to the match.  This gives:

    a) I₀ = (120 ± 1) lux

    b) Iₐ = (15 ± 2) lux

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