Question

A deviation of \[{2^ \circ }\] is produced in the yellow-ray when the prism of crown and flint glass are chromatically combined. Taking dispersive powers of crown and flint glass as 0.02 and 0.03 respectively and refractive index for yellow light for these glasses are 1.5 and 1.6 respectively. The refracting angles for the crown glass prism will be _______ ° (in degree). (Round off to the Nearest Integer).

Answer 1

Hint:In this question, we need to find the refracting angles for the crown glass prism. For this, we have to use the condition for achromatism. Whenever white light travels through a prism, it disperses into a spectrum of seven colors. However, if two prisms are joined in such a way that the total of the angular dispersions of the crown glass prism is zero, this is an achromatic prism arrangement.

Formula used:
We know that the condition for achromatism is given below.
\[{\omega _1}{\delta _1} = {\omega _2}{\delta _2}\]
Here, \[{\omega _1}\] and \[{\omega _2}\] indicate the dispersive powers of crown and flint glass prisms respectively. Also, \[{\delta _1}\] and \[{\delta _2}\] indicate the deviation for crown and flint glass prisms respectively.

The formula for deviation is given below.
\[{\delta _1} = \left( {{\mu _1} - 1} \right){A_1}\] and \[{\delta _2} = \left( {{\mu _2} - 1} \right){A_2}\]
And, \[{\mu _1}\] and \[{\mu _2}\] are the refractive indices of crown and flint glass prisms respectively. Also, \[{A_1}\] and \[{A_2}\] are the refracting angles of crown and flint glass prisms respectively.

Complete step by step solution:
According to the condition of achromatism for prism, we get
\[{\omega _1}{\delta _1} = {\omega _2}{\delta _2}\]…. (1)
Here, net deviation is \[{\delta _1} - {\delta _2} = {2^ \circ }\]…. (2)
From equation (1), we can say that
\[\dfrac{{{\omega _1}{\delta _1}}}{{{\omega _2}}} = {\delta _2}\]
Put the above value of \[{\delta _2}\] in equation (2)
Thus, we get
\[{\delta _1} - \dfrac{{{\omega _1}{\delta _1}}}{{{\omega _2}}} = {2^ \circ }\]
\[\Rightarrow {\delta _1}\left( {1 - \dfrac{{{\omega _1}}}{{{\omega _2}}}} \right) = {2^ \circ }\]

Thus, we get
\[{\delta _1}\left( {1 - \dfrac{{0.02}}{{0.03}}} \right) = {2^ \circ }\]
\[\Rightarrow {\delta _1}\left( {1 - \dfrac{2}{3}} \right) = {2^ \circ }\]
\[\Rightarrow {\delta _1}\left( {\dfrac{1}{3}} \right) = {2^ \circ }\]
By simplifying, we get
\[{\delta _1} = {6^ \circ }\]
Now, we will find the refracting angle for crown glass prism.
So, we get
\[{6^ \circ } = \left( {1.5 - 1} \right){A_1}\]
\[\Rightarrow {6^ \circ } = \left( {0.5} \right){A_1}\]
\[\Rightarrow {\dfrac{6}{{0.5}}^ \circ } = {A_1}\]
By simplifying, we get
\[{A_1} = {12^ \circ }\]

Therefore, the refracting angle for crown glass prism is \[{12^ \circ }\].

Note: Here, students generally make mistakes in finding the condition for achromatism. So, they may get a false result. Also, the calculation part is also important in this problem.