Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
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A Pin of Length 2.0 Cm Lies Along the Principal Axis of a Converging Lens, the Centre Being at a Distance of 11 Cm from the Lens. - Physics

Sum

A pin of length 2.0 cm lies along the principal axis of a converging lens, the centre being at a distance of 11 cm from the lens. The focal length of the lens is 6 cm. Find the size of the image.

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Solution

Given,
Length of the pin = 2.0 cm
Focal length (f) of the lens = 6 cm
As per the question, the centre of the pin is 11 cm away from the lens.
i.e., the object distance (u) = 10 cm

Since, we have to calculate the image of A and B, Let the image be A' and B'
So, the length of the A'B' = size of the image.
Using lens formula: \[\frac{1}{v_A} - \frac{1}{u_A} = \frac{1}{f}\]
Where vA and uA are the image and object distances from point A. 
\[\Rightarrow   \frac{1}{v_A} - \frac{1}{- 10} = \frac{1}{6}\]
\[\frac{1}{v_A} = \frac{1}{6} - \frac{1}{10} = \frac{1}{15}\]
\[v_A  = 15  \text{ cm }\]
Similarly for point B,
Lens formula: \[\frac{1}{v_B} - \frac{1}{u_B} = \frac{1}{f}\]
Where vA and uA are the image and object distances from point B.

\[\frac{1}{v_B} - \frac{1}{- 12} = \frac{1}{6}        \] 

\[ \Rightarrow  v_B  = 12  \text{ cm }\]

Length of image = vA − vB = 15 − 12 = 3 cm.

Concept: Refraction at Spherical Surfaces and by Lenses - Refraction by a Lens
  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 18 Geometrical Optics
Q 58 | Page 416
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