# A candlestick holder has a concave reflector behind the candle.The reflector magnifies a candle -0.75 times and forms an image 4.6 cm away from the reflector's surface. Is the image inverted or...

A candlestick holder has a concave reflector behind the candle.

The reflector magnifies a candle -0.75 times and forms an image 4.6 cm away from the reflector's surface. Is the image inverted or upright? What are the object distance and the reflector's focal length? Is the image virtual or real?

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The magnification of a thin lens can be represented by the formula M = -di/do = f/(f - do) where f is the focal length, di is the distance from the image to the lens, and do is the distance from the object to the lens.

Because the magnification is negative, the image is real and inverted.

do = -di/M = 4.6 cm / -0.75 = 6.13 cm

-0.75 = f/(f - 6.13) --> f = -0.75(f - 6.13) --> f(1 + 0.75) = 4.6

f = 1.75/4.6 = 0.38

We know that the real image in a concave reflector is small and inverted when magnification is -0.75. The position of object for this should be away farther than the centre of curvature and the real invetred image lies between the focus and the centre of cuvature of the reflector. Also the focal length** f** , object distance **u** and image distance **v** from the concave reflector follow the rule:

**1/f = 1/u + 1/v......................(**1). Since the magnification m =-0.75, the real image is inverted and could be caught on a screen.

The manification, **m = v/u**. Or

v/u = 0.75 Or, u = v/0.75. Substituting this value of u in (1) we get:

1/f = 1/(v/0.75)+1/v = (1/v)(1.75). So ,

the focal length, f = V/1.75 = 4.6/1.75 =** 2.62857 cm.**

U = u/0.75 = 4.6/0.75 = **6.13333 cm** is the object distance from the concave reflector.

The magnification of the image is given as -0.75 this means that the image is demagnified and inverted, and it is formed away from the reflector surface. This indicates that it is a real image. This is in line with the fact that all real images of the convex mirror are inverted.

Therefor the image is real and inverted.

Further it means that the object or the candle is situated at a point outside the center of curvature of the mirror, and that the image is situated between center of curvature and focus of the reflector. It is given that the candle is situated at distance of 4.6 cm from the mirror.

For the image magnification to be 0.75 this distance will be equal to f*(1 + 0.75).

Therefor:

4.6 = f*(1 + 0.75) = 1.75f

f = 4.6/1.75 = 2.6286

To calculate the distance of object 'do' we use the formula:

1/do = 1/f - 1/di

Where di = distance of image = 4.6 cm

Thus:

1/do = 1/2.6286 - 1/4.6 = 0.3804 - 0.2174 = 0.163

Therefor:

do = 1/0.163 = 6.135 cm