Question of the day 8
A rectangular slab of refractive index n is placed over another slab of refractive index 3, both slabs being indentical in dimensions.
If a coin is placed below the lower slab, for what value of n will the coin appear to be placed at the interface between the slabs when viewed from the top.?

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Solution

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$L e t' s a s s u m e t h i c k n e s s o f g l a s s s l a b i s t A s w e k n o w μ = \frac{R e a l d e p t h}{A p p a r e n t d e p t h} H e r e, R e a l d e p t h o f c o i n i s 2 t i n w h i c h t d i s \tan c e r a y t r a v l e i n i n r e f r a c t i v e i n d e x n a n d t d i s \tan c e r a y t r a v e l i n r e f r a c t i v e i n d e x 3 a n d a p p a r e n t d e p t h i s t D u e t o r e f r a c t i o n a t t w o b o u n d a r i e s, t h e a p p a r e n t d e p t h o f t h e c o i n w i l l b e A p p a r e n t d e p t h = \frac{R e a l d e p t h o f g l a s s s l a b o f r e f r a c t i v e i n d e x n}{μ_{n}} + \frac{R e a l d e p t h o f g l a s s s l a b o f r e f r a c t i v e i n d e x 3}{μ_{3}} A p p a r e n t d e p t h = \frac{R e a l d e p t h o f g l a s s s l a b o f r e f r a c t i v e i n d e x n}{n} + \frac{R e a l d e p t h o f g l a s s s l a b o f r e f r a c t i v e i n d e x 3}{3} t = \frac{t}{n} + \frac{t}{3} 1 = \frac{1}{n} + \frac{1}{3} \frac{1}{n} = 1 - \frac{1}{3} = \frac{2}{3} n = \frac{3}{2} = 1.5$

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