Question

An object of height h is placed midway between f and 2f in front of a biconvex lens. A real inverted image is captured on a screen placed a little beyond 2f on the other side of the lens. If the whole arrangement is immersed in water without disturbing the object, lens and the screen positions, then the new image formed will be

• A. Diminished real inverted image on the other side of the lens between f and 2f
• B. Magnified virtual erect image on the same side of the lens where the object is placed
• C. Magnified real inverted image on the other side of the lens beyond 2f
• D. Magnified real inverted image on the same side of the lens where the object is placed

Answer 1

Answer: (B)

Magnified virtual erect image on the same side of the lens where the object is placed

Answer 2

The problem describes an initial setup in air and then asks about the image formed when the entire arrangement is immersed in water.

1. Initial Setup (in Air):

• Object is placed midway between $f$ and $2f$. Let the focal length in air be $f_a$. So, the object distance $u = -1.5 f_a$.
• Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$: $\frac{1}{v} - \frac{1}{-1.5 f_a} = \frac{1}{f_a}$ $\frac{1}{v} + \frac{1}{1.5 f_a} = \frac{1}{f_a}$ $\frac{1}{v} = \frac{1}{f_a} - \frac{1}{1.5 f_a} = \frac{1}{f_a} - \frac{2}{3 f_a} = \frac{3-2}{3 f_a} = \frac{1}{3 f_a}$ $v = 3 f_a$.
• The image is formed at $3f_a$, which is beyond $2f_a$. This is consistent with the problem statement that a real, inverted image is captured beyond $2f$.
• Magnification $m = \frac{v}{u} = \frac{3 f_a}{-1.5 f_a} = -2$. The image is real, inverted, and magnified.

2. Setup Immersed in Water:

• The object and lens positions are not disturbed, so the object distance $u = -1.5 f_a$ remains the same.
• The focal length of a lens changes when it is immersed in a different medium. The lens maker's formula is $\frac{1}{f} = (\mu_{rel} - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
• Let $\mu_g$ be the refractive index of the lens material (glass) and $\mu_w$ be the refractive index of water. Assume typical values: $\mu_g = 1.5$ and $\mu_w = 4/3 \approx 1.33$.
• In air: $\frac{1}{f_a} = (\mu_g - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
• In water: $\frac{1}{f_w} = \left(\frac{\mu_g}{\mu_w} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$.
• Taking the ratio: $\frac{f_w}{f_a} = \frac{(\mu_g - 1)}{\left(\frac{\mu_g}{\mu_w} - 1\right)}$ $\frac{f_w}{f_a} = \frac{(1.5 - 1)}{\left(\frac{1.5}{4/3} - 1\right)} = \frac{0.5}{\left(\frac{3/2}{4/3} - 1\right)} = \frac{0.5}{\left(\frac{9}{8} - 1\right)} = \frac{0.5}{1/8} = 0.5 \times 8 = 4$.
• So, the new focal length in water is $f_w = 4 f_a$. The focal length of the lens increases significantly.

3. New Image Formation:

• The object distance is $u = -1.5 f_a$.

• The new focal length is $f_w = 4 f_a$.

• Compare the object distance with the new focal length: $|u| = 1.5 f_a$, and $f_w = 4 f_a$.

• Since $|u| < f_w$ (i.e., $1.5 f_a < 4 f_a$), the object is now placed within the focal length of the converging lens when it's in water.

• When an object is placed within the focal length of a converging lens, the image formed is virtual, erect, and magnified, and it is formed on the same side of the lens as the object.

• Let's verify using the lens formula: $\frac{1}{v'} - \frac{1}{u} = \frac{1}{f_w}$ $\frac{1}{v'} - \frac{1}{-1.5 f_a} = \frac{1}{4 f_a}$ $\frac{1}{v'} + \frac{1}{1.5 f_a} = \frac{1}{4 f_a}$ $\frac{1}{v'} = \frac{1}{4 f_a} - \frac{1}{1.5 f_a} = \frac{1}{4 f_a} - \frac{2}{3 f_a}$ To find a common denominator (12 $f_a$): $\frac{1}{v'} = \frac{3}{12 f_a} - \frac{8}{12 f_a} = \frac{-5}{12 f_a}$ $v' = -\frac{12}{5} f_a = -2.4 f_a$.

• Since $v'$ is negative, the image is virtual and on the same side as the object.

• New magnification $m' = \frac{v'}{u} = \frac{-2.4 f_a}{-1.5 f_a} = \frac{2.4}{1.5} = 1.6$.

• Since $m' > 1$ and positive, the image is magnified and erect.

Therefore, the new image formed will be a magnified virtual erect image on the same side of the lens where the object is placed.