Q.40. A thin convex lens and a thin concave lens are kept in contact and are co-axial. Which of the

Q: What is the correct answer for JEE Main Physics Q40?

The correct answer is A. For two thin lenses in contact, the equivalent focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_{convex}} + \frac{1}{f_{concave}}$. Since $f_{convex}$ is positive and $f_{concave}$ is negative, let $f_{convex} = f_1$ and $f_{concave} = -f_2$ (where $f_2$ is the magnitude of the concave lens's focal length). So, $\frac{1}{F} = \frac{1}{f_1} - \frac{1}{f_2} = \frac{f_2 - f_1}{f_1 f_2}$. If $f_

Q: Q.40. 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?

The correct answer is A. For two thin lenses in contact, the equivalent focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_{convex}} + \frac{1}{f_{concave}}$. Since $f_{convex}$ is positive and $f_{concave}$ is negative, let $f_{convex} = f_1$ and $f_{concave} = -f_2$ (where $f_2$ is the magnitude of the concave lens's focal length). So, $\frac{1}{F} = \frac{1}{f_1} - \frac{1}{f_2} = \frac{f_2 - f_1}{f_1 f_2}$. If $f_

Q: What are the options for this JEE Main Physics question?

A. behaves as concave lens if $f_{convex} > f_{concave}$ · B. behaves as concave lens if $|f_{convex}| f_{concave}$ · D. Focal length of the lens system will change if the position of two lenses are interchanged

Question

Q.40. 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?

FirstInTest · Accepted Answer

For two thin lenses in contact, the equivalent focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_{convex}} + \frac{1}{f_{concave}}$. Since $f_{convex}$ is positive and $f_{concave}$ is negative, let $f_{convex} = f_1$ and $f_{concave} = -f_2$ (where $f_2$ is the magnitude of the concave lens's focal length). So, $\frac{1}{F} = \frac{1}{f_1} - \frac{1}{f_2} = \frac{f_2 - f_1}{f_1 f_2}$. If $f_1 > f_2$ (i.e., $|f_{convex}| > |f_{concave}|$), then $f_2 - f_1$ is negative, so $F$ is negative. A negative focal length means the combination behaves as a concave lens. This corresponds to option (1) which states "behaves as concave lens if $f_{convex} > f_{concave}$" (assuming $f_{concave}$ here refers to its magnitude). If $f_{convex}$ and $f_{concave}$ are used as signed values, then $f_{convex} > |f_{concave}|$ would mean $f_1 > f_2$. So, option (1) is correct. Let's check other options: (2) behaves as concave lens if $|f_{convex}| < |f_{concave}|$: If $f_1 < f_2$, then $f_2 - f_1$ is positive, so $F$ is positive. The combination behaves as a convex lens. So (2) is incorrect. (3) behaves as convex lens if $f_{convex} > f_{concave}$: If $f_1 > f_2$, then $F$ is negative, so it behaves as a concave lens. So (3) is incorrect. (4) Focal length of the lens system will change if the position of two lenses are interchanged: For thin lenses in contact, the order does not matter, so the focal length will not change. So (4) is incorrect.

Answer

behaves as concave lens if $f_{convex} > f_{concave}$

Answer

behaves as concave lens if $|f_{convex}| < |f_{concave}|$

Answer

behaves as convex lens if $f_{convex} > f_{concave}$

Answer

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

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