A tank holds a layer of oil, of thickness T_o = 1.65 m. that floats on a layer o

Question

A tank holds a layer of oil, of thickness T_o = 1.65 m. that floats on a layer of syrup of thickness T_s = 0.830 m, as shown in the figure. Both liquids are dear and do not mix together. A light ray, originating at the bottom of the tank at point P, crosses the oil-syrup interface at a point 0.900 m from the axis. The ray continues and arrives at the oil-air interface, 2.00 m to the right of P and at the critical angle. What is the index of refraction of the syrup? 1.36 1.75 1.94 1.54 1.25 An object is placed 60 cm from a convex lens with a focal length of magnitude 10 cm. What is the magnification? 0.10 0.20 0.15 -0.10 -0.20

Explanation / Answer

12)

The Horizontal displacement = 2.0m - 0.9m = 1.1m

Vertical displacement of the thickness of the syrup, which is 0.83 m.
The angle of this beam is:
Arctan( 1.1 m / 0.83m )
= 52.9570
This is the critical angle for the syrup/air interface. c = 52.9570.
c = Arcsin( (index of air) / (index of syrup) )
c = Arcsin( 1 / (index of syrup) )
sinc = sin[ Arcsin( 1 / (index of syrup) ) ]
sinc = 1 / (index of syrup)
sin(52.9570) = 1 / (index of syrup)
0.79818 = 1 / (index of syrup)
index of syrup = 1 / 0.79818
index of syrup = 1.2529

option is E

13)

1/f=1/di+1/d0

1/di=1/f-1/d0

      =1/10 cm-1/60 cm

di=12 cm

magnefication, m=-di/d0

m=-12 cm/60 cm

     =-0.2

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