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# A point object is placed on the principal axis of a convex lens (f = 15 cm) at a distance of 30 cm from it. A glass plate (μ = 1.50) of thickness 1 cm is placed on the other side of the lens perpendicular to the axis. Locate the image of the point object.

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Solution

## Given, Convex lens of focal length (f) = 15 cm Object distance, (u) = −30 cm Using lens formula, $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{v}=\frac{1}{15}-\frac{1}{30}=\frac{1}{30}\phantom{\rule{0ex}{0ex}}⇒v=30$ Thus ,v = 30 cm Therefore, the image of the object will be formed at a distance of 30 cm to the right side of the lens. [since, μg = 1.5 and thickness (t) = 1 cm] $\mathrm{\Delta }t=\left(1-\frac{1}{{\mathrm{\mu }}_{\mathrm{g}}}\right)t$ $=1-\frac{2}{3}=\frac{1}{3}=0.33\mathrm{cm}$ Hence, the image of the object will be formed at 30 + 0.33 = 30.33 cm from the lens on the right side, due to the glass having thickness 1 cm.

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