Question

An object of 2 cm is placed at a distance of 50 cm from a concave mirror of focal length 15 cm . Find the position,nature and height of the image.

Answer 1

Explanation:

h1 = 2cm

u = -50cm

f = -15 cm

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ \frac{1}{ - 15} + \frac{1}{50} = \frac{1}{v} \\ \frac{35}{-750} = \frac{1}{v} \\ v = -21.42$

$magnification = \frac{ - v}{u} = \frac{+ 21.42}{ - 50} = -0.42 \\ \\ also \: \\ magnification = \frac{h2}{h1} \\ -0.42 = \frac{h2}{2} \\ h2 = -0.84 \\since \: v \: is \: -ve \: image \: is \: real \: and \: inverted\:$

Answer 2

Answer:

Position = - 21.42 cm

Nature = real and inverted

Height of the image = - 0.42 cm

Explanation:

Given that,

Height of the object (h) = 2 cm

Object distance (u) = - 50 cm (as it is always negative for concave mirror and convex lens)

Focal length (f) = - 15 cm (as it is always negative for concave mirror and convex lens)

$\mathfrak{\purple{By\:using\:mirror\:formula}}$

$\large{\frac{1}{f} = \frac{1}{v} + \frac{1}{u}}$

$\frac{1}{-15} = \frac{1}{v} + \frac{1}{-50}$

$\frac{1}{-15} - (\frac{1}{-50}) = \frac{1}{v}$

$\frac{1}{-15} + \frac{1}{50} = \frac{1}{v}$

$\frac{50 - 15}{750} = \frac{1}{v}$

$\frac{35}{-750} = \frac{1}{v}$

$\frac{-750}{35} = v$

$\boxed{\pink{\mathbb{V = - 21.42\:cm}}}$

Here, the image distance is - 21.42.

Now, we shall find out its height.

$\mathsf{\pink{By\:using\:magnification\:formula,}}$

$\frac{h'}{h} = \frac{-v}{u}$

$\frac{h'}{h} = \frac{-(-21.42)}{50}$

$\boxed{\bold{\green{h' = - 0.42}}}$

The negative sign indicates that the image is real and inverted.

Position of the image = - 21.42 cm

$\Large{\boxed{\red{Note:}}}$

• Position of the image of concave mirror is always negative as per the new cartesian sign convention.
• Also the height of the image formed by a concave mirror is always negative as it always forms a real and inverted image.