The focal length of a concave mirror is 30cm. Find the position of the object infront of the mirror so that the virtual image is three times the size of the object.Please answer.... its...

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Since the virtual image is 3 times the height of the size of the object, then `h_i=3h_0` , where `h_i` is the image height and `h_o` is the object's height.

Base on this, the magnification factor (m) of the mirror is:

`m=h_i / h_o = (3h_o)/h_o=3`

Moreover, magnification factor is equal to the negative ratio of image distance (`d_i` )  and object's distance (`d_o` ) from the mirror.

`m=-d_i / d_o`

`3=-d_i /d_o`

Expressing `d_i` in terms of `d_o` result to:

`d_i=-3d_o`

Then, apply the mirror equation to solve for `d_o` .

`1/d_o + 1/d_i = 1/f`

Substitute `d_i=-3d_o`  and f=30 cm to the formula.

`1/d_o + 1/(-3d_o) = 1/30`

`1/d_o - 1/(3d_o)=1/30`

To simplify the equation, multiply both sides by the LCD which is 30d_o.

`30d_o*(1/d_o - 1/(3d_o))=1/30*30d_o`

`30 - 10=d_o`

`20 =d_o`

Hence, the object is placed at a distance of 20 cm from the concave mirror.

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