JEE Challenger
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Combination of Coaxial Thin Convex and Concave Lenses

A thin convex lens and a thin concave lens are kept in contact and are co-axial. Which of the following statements is correct for this combination of two lenses ?

Options

A

behaves as concave lens if fconvex>fconcave|f_{\text{convex}}| > |f_{\text{concave}}|

Correct
B

behaves as concave lens if fconvex<fconcave|f_{\text{convex}}| < |f_{\text{concave}}|

C

behaves as convex lens if fconvex>fconcave|f_{\text{convex}}| > |f_{\text{concave}}|

D

Focal length of the lens system will change if the positions of two lenses are interchanged

Step-by-Step Solution

To determine the nature and focal length of the combination of a coaxial thin convex lens and a thin concave lens placed in contact, we use the effective focal length formula for thin lenses in contact:

1F=1f1+1f2\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}

where FF is the equivalent focal length of the lens combination, f1f_1 is the focal length of the convex lens, and f2f_2 is the focal length of the concave lens.

Applying the sign convention for focal lengths:

  • For a convex lens, the focal length is positive: f1=+fconvexf_1 = +|f_{\text{convex}}|
  • For a concave lens, the focal length is negative: f2=fconcavef_2 = -|f_{\text{concave}}|

Substituting these values into the expression for effective power/focal length:

1F=1fconvex1fconcave\frac{1}{F} = \frac{1}{|f_{\text{convex}}|} - \frac{1}{|f_{\text{concave}}|}

1F=fconcavefconvexfconvexfconcave\frac{1}{F} = \frac{|f_{\text{concave}}| - |f_{\text{convex}}|}{|f_{\text{convex}}| \cdot |f_{\text{concave}}|}

Now, let us analyze the behavior of the combination based on the sign of the equivalent focal length FF:

  1. Behavior as a Concave Lens: The system behaves as a net concave (diverging) lens if the equivalent focal length FF is negative (1F<0\frac{1}{F} < 0): fconcavefconvexfconvexfconcave<0\frac{|f_{\text{concave}}| - |f_{\text{convex}}|}{|f_{\text{convex}}| \cdot |f_{\text{concave}}|} < 0

    Since the denominator fconvexfconcave|f_{\text{convex}}| \cdot |f_{\text{concave}}| is always positive: fconcavefconvex<0    fconvex>fconcave|f_{\text{concave}}| - |f_{\text{convex}}| < 0 \implies |f_{\text{convex}}| > |f_{\text{concave}}|

  2. Behavior as a Convex Lens: The system behaves as a net convex (converging) lens if the equivalent focal length FF is positive (1F>0\frac{1}{F} > 0): fconcavefconvex>0    fconvex<fconcave|f_{\text{concave}}| - |f_{\text{convex}}| > 0 \implies |f_{\text{convex}}| < |f_{\text{concave}}|

  3. Effect of Interchanging Positions: Since addition is commutative (1f1+1f2=1f2+1f1\frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{f_2} + \frac{1}{f_1}), interchanging the positions of two thin lenses in contact does not change the effective focal length of the system.

Thus, the combination behaves as a concave lens if fconvex>fconcave|f_{\text{convex}}| > |f_{\text{concave}}|.

Correct Option: A

Combination of Coaxial Thin Convex and Concave Lenses | Physics PYQ Solution - JEE Challenger