1) (i) Closest distance was sqrt 2.57^2 +1^2 x 1.5x10 11 = 4x10 11(ii) Show that theta = 3.12x10 -6 radians(iii)Draw a Casegrain telescope (the correct diagram is shown in the June2011Mark scheme)(iv)The crater can be resolved in detail because the angle subtended is larger thanthe theoretical min. resolving limit2) (i) ~Apparent magnitude graph shows the variation in apparent magnitude of theeclipsing binaries.~The deeper troughs show that the less luminous star is eclipsing the moreluminous star and visa versa for the shallower troughs.~The maximum height of the graph occurs when the two star are equidistant fromthe earth ergo one quarter further in there orbit from an eclipse.~Period can be calculated as the time between two consecutive troughs of equaldepth. Therefore 5-1=4 days.~Wavelength analysis graph shows the variation in wavelength of the Balmer linecentered on ~628 nm. ~Change in wavelength is a consequence of the Doppler effect when the star ismoving towards or away from the earth in its orbit.~Red shift occurs when the star is moving away us therefore an increase in thewavelength of the light and visa versa for blue shift.~Orbital speed may be calculated by z=delta.lamda/lamda and then feeding thatanswer back into z=v/c.(ii) m - M = 5log25/10 = -0.00973)(i) T = 16111K(ii) The star was a white dwarf (iii) H balmer lines because the star's atmosphere was hot enough to havehydrogen in it's n=2 state andthe deep prominent absorption lines in the Balmerseries in the absorption spectra of stars with this surface temperature.4)(i) A black hole is an object whose escape velocity is greater than c(ii) Schwarzschild Radius, can't remember.(iii) 4.96x10 17 s