We use the formula for a camera to relate the size of the image, the size of the landscape, the height of the camera, and the focal length: \[ \frac{\text{Size of image}}{\text{Size of landscape}} = \frac{\text{Focal length}}{\text{Height of camera}}. \] Here, the size of the image is \( 2 \times 2 \) cm (so the area is 4 cm\(^2\)), the size of the landscape is 400 km\(^2\), and the height of the camera is 18 km. Substituting these values into the equation and solving for the focal length, we find that the focal length is \( 0.9 \, \text{cm} \).
Final Answer: \( 0.9 \, \text{cm} \).
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)