Q- A rectangular block of thickness 10 cm and refractive index 1-5 is placed on a small coin- A beaker is placed on the block and water - mu- - 4-3- nbsp-of height 10 cm is poured in the beaker- What will be the apparent position of the coin when viewed normally - from above -

Dear student, #160;thickness #160;of #160;the #160;glass #160;d=10cm,RI #160;of #160;glass=1.5,RI #8201;of #160;water=4/3beaker #160;depth ...

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Standard XII

Physics

Multiplication & Division of Errors

Q. A rectangu...

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Q. A rectangular block of thickness 10 cm and refractive index 1.5 is placed on a small coin. A beaker is placed on the block and water (μ = 4/3) of height 10 cm is poured in the beaker. What will be the apparent position of the coin when viewed normally , from above ?

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Solution

Dear student,

$t h i c k n e s s o f t h e g l a s s d = 10 c m, R I o f g l a s s = 1.5, R I o f w a t e r = 4 / 3 b e a k e r d e p t h = 10 c m . a p p a r e n t d e p t h b y t h e g l a s s = \frac{10}{1.5} = 20 / 3 c m a p p a r e n t d e p t h b y t h e b e a k e r = \frac{10}{4 / 3} = 15 / 2 c m t o t a l a p p a r e n t d e p t h = 20 / 3 + 15 / 2 = 14.17 c m R e g a r d s$

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Q. A rectangular block of thickness 10 cm and refractive index 1.5 is placed on a small coin. A beaker is placed on the block and water (μ = 4/3) of height 10 cm is poured in the beaker. What will be the apparent position of the coin when viewed normally , from above ?

Dear student, thickness of the glass d=10cm,RI of glass=1.5,RI of water=4/3beaker depth=10cm.apparent depth by the glass=101.5=20/3 cmapparent depth by the beaker=104/3=15/2 cmtotal apparent depth=20/3+15/2=14.17 cm Regards