The magnitudes of power of a biconvex lens (refractive index 1.5) and that of a plano-concave lens (refractive index = 1.7) are same. If the curvature of plano-concave lens exactly matches with the curvature of back surface of the biconvex lens, then ratio of radius of curvature of front and back surface of the biconvex lens is ____.

Related Formula

Core Logic

Use the Lens Maker's formula for the lens in air, and then for the lens in water. Divide the two equations to eliminate the radius terms and solve for the focal length in water.

Step 1: Lens in Air

Step 2: Lens in Water

Step 3: Ratio and Difference

Difference in focal length:

Therefore, α = 3$\alpha = 3$.

Pattern Recognition

Standard result to memorize: A glass lens (μ = 1.5$\mu = 1.5$) immersed in water (μ = 4/3$\mu = 4/3$) has its focal length become exactly 4 times its focal length in air.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Related Formula

Core Logic

The biconvex lens acts as a combination of two thin lenses in contact. For lens 1 (left): μ₁ = 1.5$\mu_1 = 1.5$, R₁ = 15 cm$R_{1} = 15 \text{ cm}$, flat side radius R₂ = ∞$R_2 = \infty$. For lens 2 (right): μ₂ = 1.2$\mu_2 = 1.2$, flat side radius R₁ = ∞$R_1 = \infty$, curved side radius R₂ = -12 cm$R_2 = -12 \text{ cm}$ (applying sign convention).

Step 1: Calculate Focal Lengths

Step 2: Calculate Equivalent Focal Length

So, fₙₑₜ = +20 cm$f_{\text{net}} = +20 \text{ cm}$.

Step 3: Image Distance and Magnification

Given object distance u = -30 cm$u = -30 \text{ cm}$.

Magnification m = (v)/(u) = (60)/(-30) = -2$m = \frac{v}{u} = \frac{60}{-30} = -2$.

Pattern Recognition

Combination of thin lenses in contact: simply add their powers 1/fₙₑₜ = 1/f₁ + 1/f₂$1/f_{net} = 1/f_1 + 1/f_2$. Apply lens maker's formula carefully with correct sign conventions for the planar boundary (R=∞$R=\infty$).

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Related Formula

Core Logic

Given δ = A$\delta_{\min} = A$.

Since A > 0$A > 0$, (A/2) < 1$\cos(A/2) < 1$. Thus, n < 2$n < 2$.

Step 1: Applying Physical Bounds

At minimum deviation, i = δ + A2$i = \frac{\delta_{\min} + A}{2}$. Since δ = A$\delta_{\min} = A$, we have i = (2A)/(2) = A$i = \frac{2A}{2} = A$. For light to enter the prism, the angle of incidence must be less than 90^°$90^\circ$ (grazing incidence is the limit). Thus, i < 90^° ⇒ A < 90^°$i < 90^\circ \Rightarrow A < 90^\circ$.

Step 2: Calculate Limits for n

Since A < 90^°$A < 90^\circ$, we have A/2 < 45^°$A/2 < 45^\circ$. The cosine function is strictly decreasing in the first quadrant, so: If A/2 < 45^°$A/2 < 45^\circ$, then (A/2) > (45^°) = 1√(2)$\cos(A/2) > \cos(45^\circ) = \frac{1}{\sqrt{2}}$.

Therefore:

Combining the limits:

Pattern Recognition

When δmin = A$\delta_{min} = A$, μ = 2 (A/2)$\mu = 2\cos(A/2)$. The extreme bounding conditions arise from A > 0$A > 0$ and i < 90^°$i < 90^\circ$. This guarantees the theoretical range √(2) < μ < 2$\sqrt{2} < \mu < 2$.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Related Formula

• Mirror focal length:

• Magnification (image size relative to object size):

where u$u$ is the object distance.

Core Logic

Given parameters:

• Radius of curvature R = 12 cm |f| = 6 cm$R = 12 \ \mathrm{cm} \implies |f| = 6 \ \mathrm{cm}$
• Object distance u = -18 cm$u = -18 \ \mathrm{cm}$

Let's calculate magnification for both mirrors:

• For Convex Mirror:
• Focal length fconvex = +6 cm$f_{\text{convex}} = +6 \ \mathrm{cm}$ (using Cartesian sign convention)
• Magnification:

Thus, image size is (1)/(4) hₒ$\frac{1}{4} h_o$.

Step 1: Calculate magnification of concave mirror

• For Concave Mirror:
• Focal length fconcave = -6 cm$f_{\text{concave}} = -6 \ \mathrm{cm}$
• Magnification:

Thus, image size is (1)/(2) hₒ$\frac{1}{2} h_o$.

Step 2: Calculate the ratio of image sizes

Since the objects are identical (same height hₒ$h_o$), the ratio of the sizes of the images is:

Thus, the ratio of sizes is 1/2$1/2$.

Pattern Recognition

Sees: Parallel convex vs concave mirror magnification. Trap: Reversing the signs of focal lengths (Convex focal length is +$+$, Concave is -$-$ in standard coordinate systems). Shortcut: Use direct magnification equation m = (f)/(f-u)$m = \frac{f}{f-u}$. For convex, m = (6)/(24) = (1)/(4)$m = \frac{6}{24} = \frac{1}{4}$. For concave, m = (-6)/(12) = -(1)/(2)$m = \frac{-6}{12} = -\frac{1}{2}$. Ratio of absolute values is (1/4)/(1/2) = 1/2$\frac{1/4}{1/2} = 1/2$.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments