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A ray of light makes a glancing angle of incidence of 60° with a flat glass surface. Given that the critical angle for glass is 42°, determine the angle of refraction r.

Physics

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Step-by-step solution

9 steps

Part 1· 4 steps

Angle of refraction

Speed of light in air: c=3×108c = 3 \times 10^{8} m/s

Speed of light in glass: vg=2×108v_g = 2 \times 10^{8} m/s

  1. Refractive index of glass

    ng=cvgn_g = \frac{c}{v_g}

    Substitute values:

    ng=3×1082×108=1.5n_g = \frac{3 \times 10^{8}}{2 \times 10^{8}} = 1.5
  2. Snell's law

    nairsin⁡i=ngsin⁡rn_{\text{air}} \sin i = n_g \sin r

    Substitute nair=1n_{\text{air}} = 1, i=45∘i = 45^\circ:

    1⋅sin⁡45∘=1.5sin⁡r1 \cdot \sin 45^\circ = 1.5 \sin r sin⁡r=sin⁡45∘1.5\sin r = \frac{\sin 45^\circ}{1.5}
  3. Compute sin⁡45∘=22\sin 45^\circ = \frac{\sqrt{2}}{2}

    sin⁡r=221.5=22×23=23\sin r = \frac{\frac{\sqrt{2}}{2}}{1.5} = \frac{\sqrt{2}}{2} \times \frac{2}{3} = \frac{\sqrt{2}}{3} sin⁡r=1.4143=0.4713\sin r = \frac{1.414}{3} = 0.4713
  4. Angle of refraction

    r=sin⁡−1(0.4713)≈28∘r = \sin^{-1}(0.4713) \approx 28^\circ

    Final answer: r=28∘r = 28^\circ

Part 2· 5 steps

Magnifying power of microscope

Focal length of objective: fo=0.50f_o = 0.50 cm
Focal length of eyepiece: fe=2.5f_e = 2.5 cm
Tube length: L=16.0L = 16.0 cm
Least distance of distinct vision: D=25D = 25 cm

  1. Image distance for objective (normal adjustment)

    L=vo+feL = v_o + f_e

    Substitute values:

    16.0=vo+2.516.0 = v_o + 2.5

    vo=16.0−2.5=13.5v_o = 16.0 - 2.5 = 13.5 cm

  2. Object distance for objective (lens formula)

    1vo−1uo=1fo\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o} 1uo=1fo−1vo\frac{1}{u_o} = \frac{1}{f_o} - \frac{1}{v_o}

    Substitute values:

    1uo=10.50−113.5=2−227\frac{1}{u_o} = \frac{1}{0.50} - \frac{1}{13.5} = 2 - \frac{2}{27} 1uo=5427−227=5227\frac{1}{u_o} = \frac{54}{27} - \frac{2}{27} = \frac{52}{27}

    uo=2752u_o = \frac{27}{52} cm

  3. Lateral magnification by objective

    mo=vo∣uo∣m_o = \frac{v_o}{|u_o|}

    Substitute values:

    mo=13.52752=13.5×5227m_o = \frac{13.5}{\frac{27}{52}} = 13.5 \times \frac{52}{27} 13.5=27213.5 = \frac{27}{2} mo=272×5227=522=26m_o = \frac{27}{2} \times \frac{52}{27} = \frac{52}{2} = 26
  4. Angular magnification by eyepiece (image at infinity)

    me=Dfem_e = \frac{D}{f_e}

    Substitute values:

    me=252.5=10m_e = \frac{25}{2.5} = 10
  5. Total magnification

    m=mo×mem = m_o \times m_e

    Substitute values:

    m=26×10=260m = 26 \times 10 = 260

    Final answer: m=260m = 260

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