NEET-XII-Physics
18: Geometrical Optics
- #12μA thin converging lens is formed with one surface convex and the other plane. Does the position of image depend on whether the convex surface or the plane surface faces the object?Ans : Yes. Using the lens maker formula we can show it. We know that the formula is:
`` \frac{1}{f}=\left(\frac{{\mu }_{2}}{{\mu }_{1}}-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{{R}_{2}}\right)``
Where, f is the focal length of the thin converging lens.
μ2 and μ1 are refractive indexes of lens and air respectively.
R1 and R2 are the radius of curvature of convex and plane surfaces respectively.
Here, R2 = ∞ because of the plane surface.
Case 1
When the convex surface is facing the object, we have:
`` \frac{1}{f}=\left(\frac{{\mu }_{2}}{{\mu }_{1}}-1\right)\left(\frac{1}{{R}_{1}}-\frac{1}{\infty }\right)``
`` \frac{1}{f}=\left(\frac{{\mu }_{2}}{{\mu }_{1}}-1\right)\frac{1}{{R}_{1}}...\left(\,\mathrm{\,i\,}\right)``
Case 2
When the plane surface is facing the object, we have:
`` \frac{1}{f}=\left(\frac{{\mu }_{2}}{{\mu }_{1}}-1\right)\left(\frac{1}{\infty }-\frac{1}{{R}_{1}}\right)``
`` \frac{1}{f}=-\left(\frac{{\mu }_{2}}{{\mu }_{1}}-1\right)\frac{1}{{R}_{1}}...\left(\,\mathrm{\,ii\,}\right)``
For Case 1, the focal length is positive and for the Case 2 the focal length is negative. Thus, the image distance is different in both cases.
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