7.1 Identify the type of electromagnetic radiation that is used in the diagram above - NSC Technical Sciences - Question 7 - 2023 - Paper 1

One property of gamma rays that makes them suitable for killing cancer cells is their high penetrating ability, which allows them to pass through tissues and target tumors effectively.

To calculate the energy of a photon, we can use the formula:

$E = \frac{hc}{\lambda}$

Where:

• $E$ is the energy of the photon,
• $h$ is Planck's constant $(6.63 \times 10^{-34} \text{J s})$,
• $c$ is the speed of light $(3 \times 10^{8} \text{m/s})$,
• $\lambda$ is the wavelength of the radiation $(5 \times 10^{-12} \text{m})$.

Plugging in the values:

$E = \frac{(6.63 \times 10^{-34})(3 \times 10^{8})}{5 \times 10^{-12}} = 3.978 \times 10^{-14} \text{J}$

Lens 1 is a convex lens, and Lens 2 is also a convex lens.

The type of image formed at point B is VIRTUAL.

The image is virtual because it is formed when the refracted rays appear to converge when extended. This means that the light rays do not actually converge at point B; instead, they diverge, creating an illusion of a virtual image.

One other practical application of convex lenses is in cameras, where they help focus light to capture clear images.