Ray Optics

JEE Main level, single correct, medium

124 of 295

Question 124 of 295

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 ?

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Answer: (A) behaves as concave lens if ∣fconvex∣>∣fconcave∣|f_{\text{convex}}| \gt |f_{\text{concave}}|

The equivalent focal length FF of two thin lenses in contact is given by 1F=1f1+1f2\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2}

For a convex lens, the focal length is positive, so f1=∣fconvex∣f_1 = |f_{\text{convex}}|

For a concave lens, the focal length is negative, so f2=−∣fconcave∣f_2 = -|f_{\text{concave}}|

Substituting these values, we get 1F=1∣fconvex∣−1∣fconcave∣\dfrac{1}{F} = \dfrac{1}{|f_{\text{convex}}|} - \dfrac{1}{|f_{\text{concave}}|}

If ∣fconvex∣>∣fconcave∣|f_{\text{convex}}| \gt |f_{\text{concave}}|, then 1∣fconvex∣<1∣fconcave∣\dfrac{1}{|f_{\text{convex}}|} \lt \dfrac{1}{|f_{\text{concave}}|}

This implies 1F<0\dfrac{1}{F} \lt 0, so FF is negative.

A negative equivalent focal length indicates that the combination behaves as a diverging or concave lens.

Interchanging the positions of the lenses does not change the equivalent focal length since 1f1+1f2=1f2+1f1\dfrac{1}{f_1} + \dfrac{1}{f_2} = \dfrac{1}{f_2} + \dfrac{1}{f_1}.

Answer: behaves as concave lens if ∣fconvex∣>∣fconcave∣|f_{\text{convex}}| \gt |f_{\text{concave}}|

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