Subject: Physics
Consider two prisms, one of crown glass and refracting angle A and the other of flint glass and refracting angle A’ placed together keeping their bases opposite to each other as shown in figure. Let µv, µ and µr be the refractive indices of the crown glass for violet, mean and red colour while µv ‘, µ’ and µr’ be the corresponding values for flint glass.
When white light is incident on the crown glass prism, it is deviated as well as dispersed which fall on the flint glass prism. The deviation produced by crown glass prism for mean colour is equal and opposite to that produced by flint glass prism. So, net deviation produced by the combination for mean colour is zero.
The deviation produced by crown glass for mean colour,
$$\delta = A(\mu – 1) $$
and the deviation, net deviation should be zero.
\begin{align*} \delta + \delta ‘ &= 0 \\ \text {or,} \: A(\mu – 1) + [-A’ (\mu ‘– 1)] &= 0 \\ \text {or,} \: A(\mu – 1) &= A’(\mu ‘– 1) \\ \text {or,} \: \frac {A’}{A} &= \frac {\mu -1}{\mu ‘ -1} \dots (i) \\\end{align*}
The angular dispersion produced by crown glass prism is
\begin{align*} \: \delta _v - \delta _r &= A(\mu _v - \mu _r ) \\ \text {and by flint glass prism,} \\ \delta _v’ - \delta _r’ &= - A’(\mu _v ‘ - \mu _r ‘) \\ \text {Net angular displacement } &= (\delta _v - \delta _r) + (\delta_v’ - \delta_r’) \\ &= A(\mu _v - \mu _r ) – A’(\mu _v ‘ - \mu _r ‘)\\ &= A[(\mu _v - \mu _r ) – \frac {A’}{A} (\mu _v ‘ - \mu _r ‘)]\\ \text {Putting value of} \frac {A’}{A} \: \text {in equation} \: (i), \text {we get} \\ \\ \end{align*}
$$Net\; angular\; dispersion\; = A[(\mu_v-\mu_r)-\frac{(\mu-1)}{(\mu'-1)} (\mu'_v-\mu'_r)]$$
$$=\;A(\mu'-1)[\frac{(\mu_v-\mu_r)}{(\mu'-1)}\;-\frac{(\mu_v-\mu'_r)}{\mu'-1}]$$
$$=\;A(\mu-1)[\omega-\omega']$$
Conclusion
(\(A\mu – 1) = A’\mu ‘– 1)\)
Since (\(\mu ‘– 1) >(A\mu – 1)\); then \(A > A’\)This means that to produce no deviation, the angle of prism of crown glass should be greater than of flint glass
3,If two prisms are of same materials, then \(\mu = \mu ‘\) and \(\omega = \omega ‘\). So, A = A’ from equation (i) and net dispersion \(\delta (\omega - \omega ‘) = 0\).
$$1,\;\omega'<\omega \;will\; be\; negative. \;This\; means\; that \;the\; order \;of \;the \;colours\; in \;the \;spectrum\; due\; to \;due\; to \;combination \;is \;opposite \;to \;that\; due\; to\; crown\; glass\; prism.$$
$$2,\;From\; equation\; (i) \;we \;have,\;
( A\mu – 1) = A’\mu ‘– 1)
Since \mu ‘– 1) >(A\mu – 1)\; , then \;A > A’.\; This\; means \;that\; to\; produce\; no \;deviation,\; the \;angle \;of\; prism\; of \;crown \;glass \;should\; be \;greater\; than \;of \;flint\; glass\;.$$
3,If two prisms are of same materials, then \(\mu = \mu ‘\) and \(\omega = \omega ‘\). So, A = A’ from equation (i) and net dispersion \(\delta (\omega - \omega ‘) = 0\).
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