00:01
The focal length of a spherical mirror is half of its radius of curvature and the radius of curvature of a concave mirror is positive and it is given to be thirty point zero centimeter which can also be written as 0 .30 meter.
00:25
Substituting for r, we get f to be equal to half of 0 .300 meter.
00:32
On calculating, we get 0 .150 meter.
00:40
The mirror equation can be written as 1 over f equals 1 over the image distance plus 1 over the 1 over the image distance plus 1 over the 1 over the.
00:52
Object distance.
00:54
Rearranging for di the image distance we get 1 over f minus 1 over do the object distance the whole inverse.
01:07
The object distance do is given to be 11 .0 cm which can also be written as 0 .110 meter.
01:19
Substituting the values we get 1 over 0 .150 meter minus 1 over 0 .110 meter the whole inverse.
01:33
On calculating we get minus 0 .41 to 5 meter.
01:41
Rounding this to 3 significant figure we get minus 0 .413 meter.
01:48
So part c.
01:50
We know that the magnification can be written as minus d .r.
01:55
Over the o substituting the values we get minus of minus 0 .4125 meters divided by 0 .110 meter on calculating we get 3 .75 subart d since the magnification is positive and is greater than 1 we can say that the image must be virtual and upright.
02:42
Subpart e...