Geometrical Optics — NEET UG practice

1. Recall how a concave lens bends parallel rays. A concave (diverging) lens spreads out any ray that travels parallel to the principal axis instead of converging it.

2. Trace the ray backward. After refraction, the outgoing ray diverges, but if we extend this diverging ray backward on the same side as the incoming ray, all such backward extensions appear to meet at a single point on the object side of the lens.

3. Identify this point. This point, from which the refracted ray appears to diverge, is the principal focus on the side of the incident ray.

4. Conclude the behaviour. So the ray does not actually pass through any focus (since it diverges), it only appears to diverge from a focal point.

Hence, the correct description is that the ray appears to diverge from the first principal focus, matching option B.

1. Handle each lens one at a time. In a lens combination, the image formed by the first lens acts as the object for the second lens, so we apply the lens formula twice. The lens formula is,

$$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$$

2. Apply it to the convex lens $L_1$. From the figure, the object distance for $L_1$ is $u=-30$ cm and its focal length is $f=+10$ cm. Therefore,

$$\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{10} - \dfrac{1}{30} = \dfrac{2}{30}$$

So,

$$v = +15\text{ cm}$$

This means $L_1$ forms an image $15$ cm to its right.

3. Treat this image as the object for the concave lens $L_2$. From the figure, this point lies on the far side of $L_2$, so it behaves as a virtual object with $u'=+12$ cm, while $f'=-10$ cm for $L_2$. Using the lens formula again,

$$\dfrac{1}{v'} = \dfrac{1}{f'} + \dfrac{1}{u'} = -\dfrac{1}{10} + \dfrac{1}{12} = -\dfrac{1}{60}$$

So,

$$v' = -60\text{ cm}$$

4. Interpret the sign. A negative $v'$ means the final image forms on the same side as the incoming light for $L_2$, i.e., to its left.

Hence, the final image is formed 60 cm to the left of the concave lens, matching option C.

1. Understand what 'index correction' means. In experiments performed on an optical bench, the tip of the pointer or the optical centre of a lens may not exactly coincide with the reading shown on the scale due to a small, fixed positioning error.

2. Recognize where this correction is applied. This mismatch is corrected using what is called an index correction, and it specifically arises in optical bench experiments, such as finding the focal length of lenses, because the lens holder's index mark and the actual optical centre can be slightly offset.

3. Check the other options. The resonance tube (sound speed), meter bridge (resistance), and simple pendulum ($g$) experiments involve their own specific corrections (like end correction or measuring the effective length), but none of these are called 'index correction'.

Hence, the correct answer is option D.