Question

An object is placed 10 cm from a concave mirror. The focal length is 5 cm. Determine (a) The image distance (b) the magnification of image

Answer 1

The image distance :

1/d¡ = 1/f – 1/d₀ 

= 1/5 – 1/10

= 2/10 – 1/10

= 1/10

d¡= 10/1 = 10 cm

The distance of the imaɡe is 10 cm.

The magnification :

m = -d¡/ d₀ = -10/10 = -1

Here, 1 means that the image is the same as the object is. The minus sign shows that the image is in inverted nature. 

Note :

If the sign will be positive than the inage will be upright.

Answer 2

Given:

• Object to mirror distance(u) = 10cm
• Focal length(f) = 5cm

To Find:

• (a) The image distance(v)

• (b) The magnification of the image

Solution:

• To find the image distance we are using the mirror equation.
• The mirror equation is $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
• Since we need to find image distance u, the above equation can be rearranged as $\frac{1}{v} =\frac{1}{f}-\frac{1}{u}$
• $\frac{1}{v}=\frac{1}{5}-\frac{1}{10}$  taking LCM we get,
• $\frac{1}{v}=\frac{1}{10}$
• v = 10cm
• To find magnification, the formula is given by,
• magnification = -v/u
• magnification = -10/10 = -1
• Since the magnification is -1, it implies that the image formed is of the same size, real and inverted(due to negative sign).

(a) The image distance(v) = 10cm

(b) magnification of the image = -1