00:01
So, this question belongs to the optical instruments in which we have a real image, a real image which is of a tree which is have magnification m equals to minus 0 .085 times by telescope's primary mirror and if a tracks image is at q equals to plus 35 centimeter, then we have to determine the distance between mirror and the tree.
00:31
So recalling the magnification formula, so magnification m, it is equals to minus q by p, where p is the distance between three and the mirror.
00:41
So substituting values, we get minus 0 .085 that is equals to minus q is 35 centimeter divided by p.
00:50
So from here we get p equals to 412 centimeters.
00:54
So, this is the distance between the object tree and the mirror.
01:01
Now recalling the mirror formula, so we get 1 by f plus 1 by p, 1 by f that is equals to 1 by p plus 1 by q.
01:10
So substituting value of p and q, so we get 1 by f equals to 1 by 412 plus 1 by 35 centimeters.
01:19
So from here after solving, we get focal length f equals to 32 .26 centimeters...