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Questions
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm. Find the position, size and nature of the image formed. Also draw the ray diagram.
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Solution
Given: Object distance (u) = −25 cm
Object height (ho) = 5 cm
Focal length (f) = +10 cm
According to the lens formula,
`1/v - 1/u = 1/f`
⇒ `1/v = 1/f + 1/u`
⇒ `1/v = 1/10 - 1/25`
⇒ `1/v = 15/250`
⇒ v = `250/15`
⇒ v = 16.66 cm
The positive value of v shows that the image is formed on the other side of the lens.
Magnification (m) = `-"Image distance"/"object distance"`
⇒ m = `-v/u`
⇒ m = `16.66/25`
⇒ m = −0.66
The negative sign shows that the image is real and formed behind the lens.
Magnification (m) =`"Image height"/"Object height"`
⇒ m = H_i/H_o` = "H"_i/5`
⇒ Hi = m × Ho
⇒ Hi = −0.66 × 5
⇒ Hi = −3.3 cm
The negative value of the image height indicates that the image is inverted. The position, size, and nature of the image are shown in the following ray diagram.

