A photographic plate is positioned directly in front of a small...

Exposure is given by the product of illuminance and time, $H = E \times t$.

The illuminance $E$ produced by a small diffused source at a distance $r$ is $E = \dfrac{I}{r^2}$, where $I$ is the luminous intensity in the direction of the plate.

According to Lambert's cosine law for a diffused source, the luminous intensity at an angle $\theta$ to the normal is $I = I_0 \cos \theta$.

Initially, the plate is directly in front of the source, so \theta = 0^{}^{{}^{\circ}}.

E_1 = \dfrac{I_0 \cos 0^{}^{{}^{\circ}}}{r^2} = \dfrac{I_0}{r^2}

For a good exposure, $H = E_1 t_1 = \dfrac{I_0}{r^2} \times 12$.

When the source is rotated by 60^{}^{{}^{\circ}}, the angle between the normal and the direction of the plate becomes \theta = 60^{}^{{}^{\circ}}.

E_2 = \dfrac{I_0 \cos 60^{}^{{}^{\circ}}}{r^2} = \dfrac{I_0}{2r^2} = \dfrac{E_1}{2}

To achieve the same exposure, $E_1 t_1 = E_2 t_2$.

$E_1 \times 12 = \dfrac{E_1}{2} \times t_2$

$t_2 = 24 \text{ s}$