Question

A candle has been placed 3 cm away from a concave mirror whose radius of curvature is 24cm. Find the position, nature and size of the object.
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Answer 1

Answer:

Answer:hope this helpful pls follow me

Answer:hope this helpful pls follow meExplanation:

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cm

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v)

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cm

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2f=-12

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2f=-121/f=1/u-1/v

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2f=-121/f=1/u-1/v-1/12=-1/3-1/v

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2f=-121/f=1/u-1/v-1/12=-1/3-1/v-1/12+1/3=-1/v

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2f=-121/f=1/u-1/v-1/12=-1/3-1/v-1/12+1/3=-1/v-1/v=-1+4/12

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2f=-121/f=1/u-1/v-1/12=-1/3-1/v-1/12+1/3=-1/v-1/v=-1+4/12-1/v=3/12

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2f=-121/f=1/u-1/v-1/12=-1/3-1/v-1/12+1/3=-1/v-1/v=-1+4/12-1/v=3/12-1/v=1/4

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2f=-121/f=1/u-1/v-1/12=-1/3-1/v-1/12+1/3=-1/v-1/v=-1+4/12-1/v=3/12-1/v=1/4v=-4

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2f=-121/f=1/u-1/v-1/12=-1/3-1/v-1/12+1/3=-1/v-1/v=-1+4/12-1/v=3/12-1/v=1/4v=-4magnification (m)=-v/u

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2f=-121/f=1/u-1/v-1/12=-1/3-1/v-1/12+1/3=-1/v-1/v=-1+4/12-1/v=3/12-1/v=1/4v=-4magnification (m)=-v/um=-(-4/-3)

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2f=-121/f=1/u-1/v-1/12=-1/3-1/v-1/12+1/3=-1/v-1/v=-1+4/12-1/v=3/12-1/v=1/4v=-4magnification (m)=-v/um=-(-4/-3)m=-4/3

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2f=-121/f=1/u-1/v-1/12=-1/3-1/v-1/12+1/3=-1/v-1/v=-1+4/12-1/v=3/12-1/v=1/4v=-4magnification (m)=-v/um=-(-4/-3)m=-4/3m=height of image(hi) /height of object(ho)

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2f=-121/f=1/u-1/v-1/12=-1/3-1/v-1/12+1/3=-1/v-1/v=-1+4/12-1/v=3/12-1/v=1/4v=-4magnification (m)=-v/um=-(-4/-3)m=-4/3m=height of image(hi) /height of object(ho) hi=m*ho

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2f=-121/f=1/u-1/v-1/12=-1/3-1/v-1/12+1/3=-1/v-1/v=-1+4/12-1/v=3/12-1/v=1/4v=-4magnification (m)=-v/um=-(-4/-3)m=-4/3m=height of image(hi) /height of object(ho) hi=m*hohi=-4/3*ho

Answer:hope this helpful pls follow meExplanation:object distance(u)=-3cmimage distance (v) radius of curvature(R)=-24cmfocal length(f)=R/2f=-24/2f=-121/f=1/u-1/v-1/12=-1/3-1/v-1/12+1/3=-1/v-1/v=-1+4/12-1/v=3/12-1/v=1/4v=-4magnification (m)=-v/um=-(-4/-3)m=-4/3m=height of image(hi) /height of object(ho) hi=m*hohi=-4/3*hoHence position of image is between pole and focus point and nature of image is real and inverted and size of image is -4/3*ho