
6
16
14
24
10
Given:
Step 1: Image Formation by \( L_1 \)
For \( L_1 \), using the lens formula:
\[ \frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1} \]
Here, \( u_1 = -12 \, \text{cm} \) (object distance is negative by convention), \( f_1 = 4 \, \text{cm} \):
\[ \frac{1}{v_1} - \frac{1}{-12} = \frac{1}{4} \]
\[ \frac{1}{v_1} = \frac{1}{4} - \frac{1}{12} = \frac{3 - 1}{12} = \frac{2}{12} = \frac{1}{6} \]
Thus, the image distance \( v_1 = 6 \, \text{cm} \) (to the right of \( L_1 \)).
Step 2: Condition for Final Image at Infinity
For the final image to be at infinity, the image formed by \( L_1 \) must lie at the focal point of \( L_2 \). Therefore:
\[ x - v_1 = f_2 \]
\[ x - 6 = 10 \]
\[ x = 16 \, \text{cm} \]
Conclusion:
The required separation \( x \) between the lenses is 16 cm.
Answer: \(\boxed{B}\)
Let $f_1$ be the focal length of lens L₁ and $f_2$ be the focal length of lens L₂.
Given $f_1 = 4$ cm and $f_2 = 10$ cm. Let $u_1$ be the object distance for L₁. $u_1 = -12$ cm (since it is to the left of L₁).
Let $v_1$ be the image distance for L₁. Using the lens formula for L₁: $\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1}$ $\frac{1}{v_1} - \frac{1}{-12} = \frac{1}{4}$ $\frac{1}{v_1} = \frac{1}{4} - \frac{1}{12}$ $\frac{1}{v_1} = \frac{3-1}{12} = \frac{2}{12} = \frac{1}{6}$ $v_1 = 6$ cm.
The image formed by L₁ acts as an object for L₂. Let $u_2$ be the object distance for L₂.
The separation between the lenses is $x$ cm.
The image formed by L₁ is at a distance of 6 cm to the right of L₁. $u_2 = -(x-6)$ cm (negative as the object is to the right of L₂). The image is formed at infinity, so $v_2 = \infty$.
Using the lens formula for L₂: $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$ $\frac{1}{\infty} - \frac{1}{-(x-6)} = \frac{1}{10}$ $0 + \frac{1}{x-6} = \frac{1}{10}$ $x-6 = 10$ $x = 10+6$ $x = 16$ cm.
Final Answer: The final answer is $\boxed{B}$
A hemispherical vessel is completely filled with a liquid of refractive index \( \mu \). A small coin is kept at the lowest point \( O \) of the vessel as shown in the figure. The minimum value of the refractive index of the liquid so that a person can see the coin from point \( E \) (at the level of the vessel) is:
A transparent block A having refractive index $ \mu_2 = 1.25 $ is surrounded by another medium of refractive index $ \mu_1 = 1.0 $ as shown in figure. A light ray is incident on the flat face of the block with incident angle $ \theta $ as shown in figure. What is the maximum value of $ \theta $ for which light suffers total internal reflection at the top surface of the block ?
Lenses that are made by combining two spherical transparent surfaces are called spherical lenses. In general, there are two kinds of spherical lenses. Lenses that are made by joining two spherical surfaces that bulge outward are convex lenses, whereas lenses that are made by joining two spherical surfaces that curve inward are concave lenses.