# Please help?

A concave mirror has a focal length of 25.3 cm. The distance between an object and its image is 42.5 cm. Find (a) the object and (b) image distances, assuming that the object lies beyond the center of curvature and (c) the object and (d) image distances, assuming that the object lies between the focal point and the mirror.

### 1 Answer

- Steve4PhysicsLv 76 months ago
Q1. The object is beyond the centre of curvature and the ray diagram is like this: http://www.leydenscience.org/physics/electmag/beyo...

Note the image is real, in front of the mirror .

f = 25.3cm (positive for concave mirror)

u = object distance (positive)

v = image distance (positive as image is real, in front of the mirror)

Distance between object and image is u - v = 42.5cm so u = 42.5 + v

Mirror equation: 1/f = 1/u + 1/v

1/25.3 = 1/(42.5 + v) + 1/v

Multiply both sides of equation by 25.3v(42.5 + v)

v(42.5 + v) = 25.3v + 25.3(42.5 + v)

42.5v + v² = 25.3v + 1075.25 + 25.3v

v² – 8.1v – 1075.25 = 0

Solve the quadratic equation with the quadratic formula. I get

v = 37.1 or -29.0

Since we know v is positive this means

v = 37.1cm

u = 42.5 + v = 42.5= + 37.1 = 79.6cm

(a) 79.6cm

(b) 37.1cm

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Q2. The object is between the focal point and the mirror and the ray diagram is like this: http://www.leydenscience.org/physics/electmag/befo...

Note the image is virtual, behind the mirror.

f = 25.3cm (positive for concave mirror)

u = object distance (positive)

v = image distance (negative as image is virtual, behind the mirror)

Distance between object and image is u + (-v) = 42.5cm so u = 42.5 + v

Note this is exactly the same values and equations as for Q1. So we’ll get the same working but we’ll need the negative solution to the quadratic equation:

v = -29.0cm

u = 42.5 + (-29.0) = 13.5cm

(c) 13.5cm

(d) -29.0cm

The arithmetic is messy. So check my working.