$(a)$ $A$ converging lens forms a real and inverted image of an object at a distance of $100 \, cm$ from it. Where should an object be placed in front of the lens so that the size of the image is twice the size of the object? Also, calculate the power of the lens.
$(b)$ State the laws of refraction.

(N/A) Given: Image distance $v = 100 \, cm$. Since the image is real and inverted, magnification $m = -2$.
Using the magnification formula $m = \frac{v}{u}$, we have $-2 = \frac{100}{u}$, which gives $u = -50 \, cm$.
Thus, the object should be placed at a distance of $50 \, cm$ in front of the lens.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$, we get $\frac{1}{f} = \frac{1}{100} - \frac{1}{-50} = \frac{1+2}{100} = \frac{3}{100}$.
So, $f = \frac{100}{3} \, cm = \frac{1}{3} \, m$.
The power of the lens $P = \frac{1}{f(\text{in } meters)} = \frac{1}{1/3} = 3 \, D$.
$(b)$ The laws of refraction are:
$1$. The incident ray, the refracted ray, and the normal to the interface of two transparent media at the point of incidence all lie in the same plane.
$2$. Snell's Law: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for the light of a given color and for the given pair of media, i.e., $\frac{\sin i}{\sin r} = \text{constant}$.