An object is placed at a distance of 6 cm from a convex mirror of focal length 12 cm. Find the position and nature of the image.

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Solution

Object distance[u] = -6cm [New Cartesian Sign Convention]

Focal length[f] = 12cm

Image distance[v] = ?

[Mirror Formula]

$1 over v equals 1 over 12 minus fraction numerator 1 over denominator negative 6 end fraction$

v = = 4cm

The image is 4 cm behind the mirror.

Magnification [m] = $straight h to the power of apostrophe over straight h equals fraction numerator negative straight v over denominator straight u end fraction$

m = $fraction numerator negative 4 over denominator negative 6 end fraction$

m = 0.66

The image is virtual,erect and diminished.

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If an object 10cm high is placed at a distance of 36 cm from a concave mirror of focal length 12cm, find the position, nature and height of image.

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(1) An object of size 7 cm is placed at 25 cm in front of concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed, so that we can get a sharp and clear image. Find nature and size of the image.

(2) An object of height 5 cm is held 20 cm away from converging lens of focal length 10 cm. Find the position, nature and size of the image formed.

(3) An object is placed at a distance of 10 cm from a convex lens of focal length 12 cm. Find the position and nature of the image.

Q. An object is placed at a distance of

10

cm from a convex mirror of focal length

15

cm. Find the position and nature of the image.

Q. Question 12
An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

An object of 4cm hieght is placed at a distance 36cm from the conacve mirror having 12cm focal length. Find the position of the nature and hieght of an image.

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An object is placed at a distance of 6 cm from a convex mirror of focal length 12 cm. Find the position and nature of the image.

Object distance[u] = -6cm [New Cartesian Sign Convention] Focal length[f] = 12cm Image distance[v] = ? [Mirror Formula] v = = 4cm The image is 4 cm behind the mirror. Magnification [m] = m = m = 0.66 The image is virtual,erect and diminished.