question_answer

In figure S is a monochromatic source emitting light of wavelength \[\lambda =500\,\,nm\] and a thin lens of focal length \[0.10\,\,m\] is cut into identical halves. The two halves are placed symmetrically about the central axis SO with a gap of\[d=0.5\,mm\]. The distance of the lens from S is \[0.150\,m\] and that from O is \[1.30\,m\]. If the third intensity maximum occurs at a point A on the screen (perpendicular to SO), find the distance OA.

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.