Question

An object is placed perpendicular to the principal axis of a convex lens of focal 20 cm.the distance of the object from the lens is 24cm .find the position and nature of ​

Answer 1

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It is given that :-

• Focal length of lens = 20cm
• Object distance = 24cm.
• Image distance = v = ?

Now, we know according to lens formula :-

$\maltese \: \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$\maltese \: f = \frac{uv}{u - v}$

In this formula :-

• v = Image distance
• u = Object distance
• f = focal length of lens.

Now, after inputting the known values we get :-

$\maltese \: 20 = \frac{ - 24v}{24 - v}$

$\maltese \: 20(24 - v) = - 24v$

$\maltese \: 480 - 20v = - 24v$

$\maltese \: 480 = - 24v + 20v$

$\maltese \: 480 = 4v$

$\maltese \: v = \frac{480}{4}$

$\maltese \: v =\huge \cancel \frac{480}{4}$

$\maltese \: v = 120cm$

Therefore we get that the image will form at 120cm from the pole of lens.

Now we can see that the Object distance is greater than focal length of lens, this means object is between the Focus and Curvature, Now in this case the Image formed will be Real, Inverted and Magnified .

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BY SCIVIBHANSHU

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