JEE MAIN - Physics (2019 - 10th April Morning Slot - No. 7)
A ray of light AO in vacuum is incident on a
glass slab at angle 60° and refracted at angle 30°
along OB as shown in the figure. The optical path
length of light ray from A to B is:
_10th_April_Morning_Slot_en_7_1.png)
2a + 2b/$$\sqrt 3$$
2a + 2b/3
2a + 2b
2$$\sqrt3$$/a + 2b
Explanation
From Snell’s law
1 sin 60° = $$\mu $$ sin 30°
$$ \Rightarrow $$ $$\mu $$ = $$\sqrt 3 $$
Optical path = AO + $$\mu $$(OB)
= $${a \over {\cos {{60}^o}}} + \sqrt 3 {b \over {\cos {{30}^o}}}$$
= 2a + 2b
1 sin 60° = $$\mu $$ sin 30°
$$ \Rightarrow $$ $$\mu $$ = $$\sqrt 3 $$
Optical path = AO + $$\mu $$(OB)
= $${a \over {\cos {{60}^o}}} + \sqrt 3 {b \over {\cos {{30}^o}}}$$
= 2a + 2b
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