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Q.An object of height 1 mm is kept perpendicular to the axis of a thin convex lens of power +10 D. The distance between the object and the lens is 15 cm. Find the position and height of the image formed.

Karnataka PUCKarnataka II PUC Board 2025Subjective· 5mImportance★★★★★
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Focal length f=1/P=10f = 1/P = 10 cm. Using the lens formula with u=−15u=-15 cm gives v=+30v=+30 cm (real, on the far side). Magnification m=v/u=−2m=v/u=-2, so the image is inverted and 2 mm tall.

Given: Object height h=1 mmh = 1\,\text{mm}; power P=+10 DP = +10\,\text{D}; object distance u=−15 cmu = -15\,\text{cm} (measured against the incident light, sign convention).

Focal length:

f=1P=110 m=0.10 m=+10 cmf = \frac{1}{P} = \frac{1}{10}\,\text{m} = 0.10\,\text{m} = +10\,\text{cm}

(convex lens, positive).

Lens formula:

1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

1v=1f+1u=110+1−15=110−115\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{-15} = \frac{1}{10} - \frac{1}{15}

1v=3−230=130\frac{1}{v} = \frac{3 - 2}{30} = \frac{1}{30}

v=+30 cmv = +30\,\text{cm}

The positive sign means the image is formed 30 cm on the other side of the lens — it is real.

Magnification:

m=vu=30−15=−2m = \frac{v}{u} = \frac{30}{-15} = -2 …

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