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NEET-XII-Physics

20: dispersion and Spectra

with Solutions - page 4
Qstn# iv-11 Prvs-Qstn
  • #11
    The refractive index of a material M1 changes by 0.014 and that of another material M2 changes by 0.024 as the colour of the light is changed from red to violet. Two thin prisms, one made of M1(A = 5.3°) and the other made of M2(A = 3.7°) are combined with their refracting angles oppositely directed. (a) Find the angular dispersion produced by the combination. (b) The prisms are now combined with their refracting angles similarly directed. Find the angular dispersion produced by the combination.
    Ans : If μ'v and μ'r are the refractive indices of material M1, then we have:
    μ'v - μ'r = 0.014
    If μv and μr are the refractive indices of material M2, then we have:
    μv - μr = 0.024
    Now,
    Angle of prism for M1, A' = 5.3°
    Angle of prism for M2, A = 3.7° (a) When the prisms are oppositely directed, angular dispersion `` \left({\delta }_{1}\right)`` is given by
    δ1 = (μv - μr)A - (μ'v - μ'r)A'
    On substituting the values, we get:
    δ1 = 0.024 × 3.7° - 0.014 × 5.3°
    = 0.0146°
    So, the angular dispersion is 0.0146°. (b) When the prisms are similarly directed, angular dispersion`` \left({\delta }_{2}\right)`` is given by
    δ2 = (μv - μr)A + (μ'v - μ'r)A'
    On substituting the values, we get:
    δ2 = 0.024 × 3.7° + 0.014 × 5.3°
    = 0.163°
    So, the angular dispersion is 0.163°.
  • #11-a
    Find the angular dispersion produced by the combination.
    Ans : When the prisms are oppositely directed, angular dispersion `` \left({\delta }_{1}\right)`` is given by
    δ1 = (μv - μr)A - (μ'v - μ'r)A'
    On substituting the values, we get:
    δ1 = 0.024 × 3.7° - 0.014 × 5.3°
    = 0.0146°
    So, the angular dispersion is 0.0146°.
  • #11-b
    The prisms are now combined with their refracting angles similarly directed. Find the angular dispersion produced by the combination.
    Ans : When the prisms are similarly directed, angular dispersion`` \left({\delta }_{2}\right)`` is given by
    δ2 = (μv - μr)A + (μ'v - μ'r)A'
    On substituting the values, we get:
    δ2 = 0.024 × 3.7° + 0.014 × 5.3°
    = 0.163°
    So, the angular dispersion is 0.163°.