Question

The radius of curvature of a convex mirror is 30 cm. An object of height 5 cm is kept at a
distance of 10 cm from the pole. Find the nature, size and magnification of image.​

Answer 1

R=30cm. height of object(h)=5cm

therefore, u= -10cm('-'from sign convention)

f=R/2=15cm

by mirror formula

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

1/v=1/f-1/u=1/15+1/10=5/30

v=30/5=+6 cm

'+' sign indicates that image is formed b/w focus and pole .

magnification

$m = - \frac{v}{u} = \frac{hi}{ho}$

hi=-v×ho/u

hi=+3cm

'+' sign means that image is formed above principal axis.

m=hi/ho=3/5

positive sign of M indicates that the image is real virtual and diminished .

hope it helps

Answer 2

Given, radius of curvature of a convex mirror (R)= +30cm

Given, radius of curvature of a convex mirror (R)= +30cmtherefore, Focal length (f) = R/2 = +30/2 = +15cm.

Given, radius of curvature of a convex mirror (R)= +30cmtherefore, Focal length (f) = R/2 = +30/2 = +15cm.object distance (u) = -10cm.

Given, radius of curvature of a convex mirror (R)= +30cmtherefore, Focal length (f) = R/2 = +30/2 = +15cm.object distance (u) = -10cm.Height of object (h) = 5cm.

Given, radius of curvature of a convex mirror (R)= +30cmtherefore, Focal length (f) = R/2 = +30/2 = +15cm.object distance (u) = -10cm.Height of object (h) = 5cm.we have, 1/f = 1/v + 1/u

Given, radius of curvature of a convex mirror (R)= +30cmtherefore, Focal length (f) = R/2 = +30/2 = +15cm.object distance (u) = -10cm.Height of object (h) = 5cm.we have, 1/f = 1/v + 1/u 1/v = 1/f - 1/u = 1/15 - (-1/10)

Given, radius of curvature of a convex mirror (R)= +30cmtherefore, Focal length (f) = R/2 = +30/2 = +15cm.object distance (u) = -10cm.Height of object (h) = 5cm.we have, 1/f = 1/v + 1/u 1/v = 1/f - 1/u = 1/15 - (-1/10) 1/15 + 1/10 = 2+3/30 = 5/30

Given, radius of curvature of a convex mirror (R)= +30cmtherefore, Focal length (f) = R/2 = +30/2 = +15cm.object distance (u) = -10cm.Height of object (h) = 5cm.we have, 1/f = 1/v + 1/u 1/v = 1/f - 1/u = 1/15 - (-1/10) 1/15 + 1/10 = 2+3/30 = 5/30 1/v = 1/6

Given, radius of curvature of a convex mirror (R)= +30cmtherefore, Focal length (f) = R/2 = +30/2 = +15cm.object distance (u) = -10cm.Height of object (h) = 5cm.we have, 1/f = 1/v + 1/u 1/v = 1/f - 1/u = 1/15 - (-1/10) 1/15 + 1/10 = 2+3/30 = 5/30 1/v = 1/6 v = 6cm.

Given, radius of curvature of a convex mirror (R)= +30cmtherefore, Focal length (f) = R/2 = +30/2 = +15cm.object distance (u) = -10cm.Height of object (h) = 5cm.we have, 1/f = 1/v + 1/u 1/v = 1/f - 1/u = 1/15 - (-1/10) 1/15 + 1/10 = 2+3/30 = 5/30 1/v = 1/6 v = 6cm.Magnification, m = -v/u = h'/h

Given, radius of curvature of a convex mirror (R)= +30cmtherefore, Focal length (f) = R/2 = +30/2 = +15cm.object distance (u) = -10cm.Height of object (h) = 5cm.we have, 1/f = 1/v + 1/u 1/v = 1/f - 1/u = 1/15 - (-1/10) 1/15 + 1/10 = 2+3/30 = 5/30 1/v = 1/6 v = 6cm.Magnification, m = -v/u = h'/h -(6/-10) = h'/5

Given, radius of curvature of a convex mirror (R)= +30cmtherefore, Focal length (f) = R/2 = +30/2 = +15cm.object distance (u) = -10cm.Height of object (h) = 5cm.we have, 1/f = 1/v + 1/u 1/v = 1/f - 1/u = 1/15 - (-1/10) 1/15 + 1/10 = 2+3/30 = 5/30 1/v = 1/6 v = 6cm.Magnification, m = -v/u = h'/h -(6/-10) = h'/5 3/5 = h'/5

Given, radius of curvature of a convex mirror (R)= +30cmtherefore, Focal length (f) = R/2 = +30/2 = +15cm.object distance (u) = -10cm.Height of object (h) = 5cm.we have, 1/f = 1/v + 1/u 1/v = 1/f - 1/u = 1/15 - (-1/10) 1/15 + 1/10 = 2+3/30 = 5/30 1/v = 1/6 v = 6cm.Magnification, m = -v/u = h'/h -(6/-10) = h'/5 3/5 = h'/5 h' = 3cm. Ans.

Given, radius of curvature of a convex mirror (R)= +30cmtherefore, Focal length (f) = R/2 = +30/2 = +15cm.object distance (u) = -10cm.Height of object (h) = 5cm.we have, 1/f = 1/v + 1/u 1/v = 1/f - 1/u = 1/15 - (-1/10) 1/15 + 1/10 = 2+3/30 = 5/30 1/v = 1/6 v = 6cm.Magnification, m = -v/u = h'/h -(6/-10) = h'/5 3/5 = h'/5 h' = 3cm. Ans.image will be virtual and erect. Ans.