A) \[2h{{\left( {{\mu }^{2}}-1 \right)}^{1/2}}\]
B) \[2h{{\left( {{\mu }^{2}}-1 \right)}^{1/2}}\]
C) \[h/\left[ 2{{\left( {{\mu }^{2}}-1 \right)}^{1/2}} \right]\]
D) \[h/{{\left( {{\mu }^{2}}-1 \right)}^{1/2}}\]
Correct Answer: A
Solution :
[a] The figure shows incidence from water at critical angle \[{{\theta }_{c}}\] for the limiting case. Now, \[\sin {{\theta }_{c}}=1/\mu \] so that \[\tan {{\theta }_{c}}=1/{{\left( {{\mu }^{2}}-1 \right)}^{1/2}}\] which is also equal to r/h where r is the radius of the disc. Therefore, diameter of the disc is \[2r=2h\tan {{\theta }_{c}}\]You need to login to perform this action.
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