Date | November 2012 | Marks available | 1 | Reference code | 12N.1.SL.TZ0.27 |
Level | Standard level | Paper | Paper 1 | Time zone | Time zone 0 |
Command term | Question number | 27 | Adapted from | N/A |
Question
The intensity of radiation from a star at the surface of one of its planets is I. The distance between the star and the planet is d.
What is the intensity at the surface of another planet which is a distance \(\frac{d}{4}\) from the star?
A. 4I
B. 8I
C. 16I
D. 64I
Markscheme
C
Examiners report
This was a slightly unusual question in that the inverse square law is usually used by starting close to an object and then moving away; here it was used in reverse. Nevertheless, one distance was four times further than the other and so the intensity ratio would be 1:16 with it therefore being 16 times greater for the planet closer to the star.