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प्रश्न
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.
उत्तर १
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 = 1/10 - 1/25 = 15/250`
`v=250/15=16.66` cm
The positive value of v shows that the image is formed at the other side of the lens.
Magnification m =`-"image distance"/"object distance"=-v/u = 16.66/25 = -0.66`
The negative sign shows that the image is real and formed behind the lens.
Magnification m =`"image height"/"object height" = "H"_i/"H"_o = "H"_i/5`
Hi = m × Ho= −0.66 × 5 = −3.3 cm
The negative value of the image height indicates that the image is inverted. The position, size, and nature of image are shown in the following ray diagram.
उत्तर २
Given:
Object distance, u = −25 cm
Focal length, f = 10 cm
Height of the object, h = 5 cm
Applying the lens formula, we get:
`1/f = 1/v - 1/u`
⇒`1/v = 1/f + 1/u`
⇒`1/v = 1/f + 1/u`
⇒`11/2/v = 1/10 + 1/-25`
⇒`1/v = (5 - 2)/50`
⇒ `1/v = 3/50`
⇒ v = `50/3`
⇒ v = 16.67 cm
The image will be at a distance of v cm behind the lens.
m = `"h'"/"h" = "v"/"u"`
⇒ `"h'"/5 = (50/3)/(-25)`
⇒ `"h'"/5 = 50/(3 xx -25)`
⇒ `"h'"/5 = 2/(-3)`
⇒ −3h' = 5 × 2
⇒ −3h' = 10
⇒ h' = `10/(-3)`
⇒ h' = – 3.33 cm
Size of the image: The image is smaller than the object. And the negative sign indicates that the image is real and inverted.
Nature of image: Real and inverted.
Ray diagram:
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