A) \[0.5\,mm\]
B) \[1.0\,mm\]
C) \[1.5\,\,mm\]
D) \[2.0\,\,mm\]
Correct Answer: B
Solution :
The image of sources S formed in the two half lenses are \[d=\frac{\sqrt{14\times 6-9}}{\sqrt{6}}\,=\frac{5\sqrt{3}}{\sqrt{6}}\,=\frac{5}{\sqrt{2}}\] and \[\therefore \,N=5\] whose third order maximum is formed at point A. Now, \[(a-b){{x}^{2}}+ax+1=0\left\langle _{2\alpha }^{\alpha } \right.\] \[3\alpha =\frac{-a}{a-b};\,\,2\alpha =\,\frac{1}{a-b}\] \[\therefore \,\,\frac{2{{a}^{2}}}{{{(a-b)}^{2}}9}=\frac{1}{a-b}\,\Rightarrow 2{{a}^{2}}=9(a-b)\] \[\Rightarrow \,2{{a}^{2}}-9a+9b=0\] \[a\in R,\]= 1 mm.You need to login to perform this action.
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