Solveeit Logo

Question

Question: The refractive indices of flint glass prisms for \(C,D\) and \(F\) lines are \(1.790,1.795\) and \(1...

The refractive indices of flint glass prisms for C,DC,D and FF lines are 1.790,1.7951.790,1.795 and 1.8051.805 respectively. The dispersive power of the flint glass prism is
A)0.141 B)0.125 C)0.018 D)0.929 \begin{aligned} & A)0.141 \\\ & B)0.125 \\\ & C)0.018 \\\ & D)0.929 \\\ \end{aligned}

Explanation

Solution

Dispersive power of a prism is defined as the ratio of angular dispersion to the angle of deviation for the mean wavelength. C,DC,D and FF lines of a prism correspond to red, yellow and violet lines of the prism. Angle of deviation of a particular color of light when passed through a prism is dependent on the refractive index of that particular color of light.

Formula used:
ω=δvδrδy=μvμrμy1\omega =\dfrac{{{\delta }_{v}}-{{\delta }_{r}}}{{{\delta }_{y}}}=\dfrac{{{\mu }_{v}}-{{\mu }_{r}}}{{{\mu }_{y}}-1}

Complete step by step answer:
Dispersive power of the material of a prism refers to the ratio of angular dispersion to the angle of deviation for the mean wavelength. Here, angle of deviation for the mean wavelength corresponds to the yellow line of the prism. C,DC,D and FF lines of a prism correspond to red, yellow and violet lines of the prism. Mathematically, dispersive power of a prism is given by
ω=δvδrδy\omega =\dfrac{{{\delta }_{v}}-{{\delta }_{r}}}{{{\delta }_{y}}}
where
ω\omega is the dispersive power of a prism
δv{{\delta }_{v}} is the angle of deviation of violet light
δr{{\delta }_{r}} is the angle of deviation of red light
δy{{\delta }_{y}} is the angle of deviation of yellow light (mean wavelength)
δvδr{{\delta }_{v}}-{{\delta }_{r}} refers to the angular dispersion of the prism
Let this be equation 1.
The figure given below can be referred for clarity.

Since angular dispersion and angle of deviation for the mean wavelength are related to the respective refractive indices of red, yellow and violet lines of the prism, dispersive power of the prism is also related to the refractive indices of the mentioned. Therefore, equation 1 can be rewritten as
ω=δvδrδy=μvμrμy1\omega =\dfrac{{{\delta }_{v}}-{{\delta }_{r}}}{{{\delta }_{y}}}=\dfrac{{{\mu }_{v}}-{{\mu }_{r}}}{{{\mu }_{y}}-1}
where
μv{{\mu }_{v}} is the refractive index corresponding to violet light or FF line
μr{{\mu }_{r}} is the refractive index corresponding to red light or CC line
μy{{\mu }_{y}} is the refractive index corresponding to yellow light or DD line
Let this be equation 2.
Coming to our question, we are provided with a flint glass prism. It is also given that
μr=1.790{{\mu }_{r}}=1.790, is the refractive index corresponding to CC line of the flint glass prism
μv=1.805{{\mu }_{v}}=1.805, is the refractive index corresponding to FF line of the prism
μy=1.795{{\mu }_{y}}=1.795, is the refractive index corresponding to DD line of the prism
Substituting these values in equation 2, we have
ω=μvμrμy1=1.8051.7901.7951=0.0150.795=0.018\omega =\dfrac{{{\mu }_{v}}-{{\mu }_{r}}}{{{\mu }_{y}}-1}=\dfrac{1.805-1.790}{1.795-1}=\dfrac{0.015}{0.795}=0.018{}^\circ
Therefore, dispersive power of the given flint glass prism is equal to 0.0180.018{}^\circ . The correct answer is option CC.

Note:
The question can also be approached in a different way, using the formula given below:
ω=μFμCμD1\omega =\dfrac{{{\mu }_{F}}-{{\mu }_{C}}}{{{\mu }_{D}}-1}
where
μF{{\mu }_{F}} is the refractive index corresponding to FF line
μC{{\mu }_{C}} is the refractive index corresponding to CC line
μD{{\mu }_{D}} is the refractive index corresponding to DD line
This can be considered as a direct approach to the question and students need not think about red, yellow and violet lines, here.