convex mirror and focal length?

Question

A candle is placed 21.9 cm in front of a convex mirror. When the convex mirror is replaced with a plane mirror, the image moves 15.0 cm farther away from the mirror. Find the focal length of the convex mirror.

Answer 1

The object is placed 21.9 cm in front of the mirror.

Therefore a plane mirror will form an image 21.9 cm behind the mirror.

The image by the plane mirror is 15.0 cm farher than the image by the convex mirror.

Therefore the convex mirror forms an image at 21.9 - 15.0 = 6.9 cm behind the mirror.



Using coordinate sign convention:

Object distance u = -21.9 cm

Image distance v = 6.9 cm

1/f = 1/v + 1/u

1/f = 1/6.9 - 1/21.9

1/f = 0.0993

f = 1/0.0993 = 10.1 cm

Ans: 10.1 cm



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Note: If your school uses the other sign convention in which the distance to real is taken as positive and to imaginary is taken as negative, then make u = 21.9 and v = -6.9

Then the answer will be -10.1 cm

Answer 2

Because in plain mirror the image is at the same disatnce as object on opposite side. So Image disrtance with convex mirror is 21.9 -15.0, or = 6.9 cm; using the formula 1/v + 1/u = 1/f, we get with u = 21.9 cm and v = - 6.9 cm

1/f = -1/6.9 + 1/21.9 = 1/f = -10/69 + 10/219 or

1/f = (-2190 + 690)/(219*69) = -1500/(69*219) or f = - 10.07 or

Focal length is 10.1 cm and focus is behind teh mirror.

Answer 3

Let the Focal Length be f, and the radius of curvature R=30cm. f=R/2=30/2=15cm. Now, let the image distance be v, and the object dist. u= -20cm(minus bcoz we are assuming object to be on left side of mirror) Then, 1/v + 1/u = 1/f (By Mirror Formula) Substituting the values, 1/v + 1/(-20) = 1/(-15) (f is negative bcoz centre of curvature is assumed to be on the left side of the mirror) 1/v = 1/20 - 1/15 = (15 - 20)/15*20 = -5/15*20 = -1/60 Thus, v = -60cm. The -ve sign indicates that the image is formed to the left of the mirror, i.e., to the same side of the mirror as the object. Thus, the image is real. h(i)/h(o) = -v/u where h(i) is the height of the image and h(o) is the height of the object. h(i) = -(-60)/(-20) = -3cm The -ve sign indicates that the image is 3 cm below the principal axis. Which means that the image is inverted. Finally, magnification = h(i)/h(o) = -3/4 = -0.75. I hope this solution helps you, trust me the solution given below bu Piyush is incorrect.