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

Options

A. appear to be near A.
B. appear to be near D.
C. appear to be at the centre of AD.
D. not be seen at all.

Answer

A

Explanation

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.