Question

The convex surface of a thin concavo-convex lens of glass of refractive index $1.5$ has a radius of curvature $20cm$. The concave surface has a radius of curvature $60cm$. The convex side is silvered and placed on a horizontal surface.
If the concave part is filled with water of refractive index $4/3$, find the distance through which the pin should be moved so that the image of the pin again coincides with the pin?

• A. $\Delta x=1.15cm,up$
• B. $\Delta x=3.15cm,down$
• C. $\Delta x=1.15cm,down$
• D. $\Delta x=0.05cm,up$

Answer 1

Answer: (C)

Method 1 : When the concave part is filled with water of refractive index $4/3$, the optical arrangement is equivalent to concave mirror of focal length $F$ such that
$\frac{1}{F}=\frac{1}{f_w}+\frac{1}{f_g}+\frac{1}{f_m}+\frac{1}{f_g}+\frac{1}{f_w}$
$\frac{1}{f_w}=(\frac{4}{3}-1)(\frac{1}{60}-\frac{1}{\infty })=\frac{1}{180}$
$f_w=180cm$
$f_g=60cm$ (calculated earlier)
$\frac{1}{F}=\frac{1}{180}+\frac{1}{60}+\frac{1}{10}+\frac{1}{60}+\frac{1}{180}=\frac{26}{180}$
$F=\frac{180}{26}cm$
$X_1=R=2F=\frac{180}{26}\times 2=\frac{180}{13}=13.85cm$
Method 2 : We use the equation
$\frac{n_2}{x_2}-\frac{n_1}{x_1}=\frac{n_2-n_1}{R}$
For refraction at the interface ${}^{\prime }{1}^{\prime }$ (air water),
$\frac{4/3}{x_2}-\frac{1}{x_1}=\frac{4/3-1}{\infty }....\left(i\right)$
The image of interface ${}^{\prime }{1}^{\prime }$ is the object for the interface ${}^{\prime }{2}^{\prime }$.
$\frac{1.5}{+20}-\frac{1}{x_1}=\frac{1.5-4/3}{+60}$
$X_1=\frac{360}{26}=13.85$
$△X=15.0-13.85=1.15cm$
Method 3 : Using lensmaker's formula and the relation
$\frac{1}{F}=\frac{1}{x_2}-\frac{1}{x_1}$
$f_m=180cm$ (using lensmaker's formula)
$f_g=60cm$ (using lensmaker's formula)
$\frac{1}{-180}=\frac{1}{x_2}-\frac{1}{x_1}$ (for the water lens) ... (i)
$\frac{1}{-60}=\frac{1}{+20}-\frac{1}{x_1}$ (for the glass lens) ... (ii)
(The image by the water lens is the object for the glass lens and if the image by the glass lens is at $+20$, then the rays will fall normally on the mirror.)
Adding Eqs. (i) and (ii).
$\frac{1}{-180}+\frac{1}{-60}=\frac{1}{20}-\frac{1}{x_1}$
$x_1=\frac{180}{3}=13.85cm$
$△x=15.0-13.85=1.15cm$.