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A rectangular block of glass ABCD has a refractive index 1.6. A pin is placed midway on the face AB (See figure). When observed from the face AD, the pin shall |
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Answer» appear to be near A.  Situation given in the statement is depicted in above figure. Here if glass to air critical angle is C then, `sinC=(1)/(mu)` (Where `mu` = refractive index of glass w.r.t. air) `therefore sinC=(1)/(1.6)` `therefore` sinC=0.625 ...(1) `therefore C=38^@40.` ...(2) Suppose angle `theta` shown in the figure is SMALLER than C. From the geometry of figure,`sintheta=(AE)/(EP)` `therefore sin theta=(y)/(sqrt(y^2+(a^2)/(4)))` ....(3) (Where a = side length of a square) Here when `theta lt C` in above equation, light will come out of point E and hence object at point P can be seen. If `theta` is INCREASED then value of y will also increase. Suppose when `theta = C, y = y_0` and so from above equation, `sinC=(y_0)/(sqrt(y_0^2+(a^2)/(4)))` `therefore 0.625=(y_0)/(sqrt(y_0^2+(a^2)/(4)))` [From equation (1)] Taking square, `therefore (y_0^2)/(y_0^2+(a^2)/(4))=0.4` `therefore y_0^2=0.4y_0^2+0.4xx(a^2)/(4)` `therefore 0.6y_0^2=(a^2)/(10)` `therefore y_0^2=(a^2)/(6)` `therefore y_0=(a)/(sqrt6)=(a)/(2.449)=(a)/(2.5)` `therefore y_0=0.4 a lt 0.5 a` When we observe with our eye, nearer to END A, object at P can be seen. Hence, option (A) is correct. |
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