NEET Lens Makers Formula/ Len's Formula MCQs & PYQs

Assertion (A) is true; a lens acts as a diverging lens if the surrounding medium's refractive index is greater than the lens material's. Reason (R) is true; a concave lens (a common diverging lens) has this shape. R describes the shape of a diverging lens, not why a convex lens changes its behavior in a different medium. Thus, both are true, but R does not explain A.

Red light has the longest wavelength and scatters least, making it visible from a distance. Thus, (A) is true. However, the human eye is most sensitive to yellow-green light (approximately \(555\text{ nm}\)), not red. Thus, (R) is false. So, (A) is true but (R) is false.

A single convex lens suffers from chromatic aberration due to dispersion, so (A) is true. For monochromatic light, ideal parallel rays passing through a convex lens *do* converge at a single focal point (ignoring spherical aberration). Hence, (R) is false.

The focal length of a lens is given by \(1/f = (n-1)(1/R_1 - 1/R_2)\). Since the refractive index \(n\) varies with the color of light, focal length is different for different colors. Thus (A) is false. Focal length depends on \(n\), \(R_1\), \(R_2\), and the surrounding medium, not only the material. Also, \(n\) for a material depends on color. So (R) is false. Both (A) and (R) are false.

For a convex mirror, focal length \(f = +26 \text{ cm}\). Object distance \(u = -26 \text{ cm}\). Using mirror formula \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\) gives \(\frac{1}{v} = \frac{1}{26} - \frac{1}{-26} = \frac{2}{26} = \frac{1}{13}\) so \(v = 13 \text{ cm}\). (A) is false. The correct mirror formula is \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\) not \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\). (R) is false. Thus, both (A) and (R) are false.

A biconvex lens can form a virtual image of a virtual object, for instance, when intercepting converging rays. So (A) is true. The nature of a lens (converging or diverging) is determined by the refractive index of its material relative to the surrounding medium. If \(\mu_{lens} > \mu_{medium}\), a biconvex lens converges; otherwise, it diverges. So (R) is true and explains (A).

Assertion (A) is true. A real object producing an erect, magnified image (like (2) times) can only happen with a convergent (convex) lens when the object is between (F) and (O).
Reason (R) is true. Concave lenses produce erect, diminished images; convex lenses produce erect, magnified images under specific conditions.
Both (A) and (R) are true, but (R) does not explain the magnification condition in (A).

Assertion (A) is true. (m = +0.5) indicates an erect and diminished image. For a real object, only a divergent (concave) lens produces such an image.
Reason (R) is true. A concave lens always forms a virtual, erect, and diminished image for a real object.
(R) correctly explains (A) because a concave lens's image characteristics match the given magnification.

To get a real image of the same size, the image distance from the lens must be \(v = 2f\). Here, \(v = x\), therefore \(x = 2f ⇒ f = \frac{x}{2}\).