Distance between object and its image (magnified by -(1)/(3)$-\frac{1}{3}$) is 30 cm$30\text{ cm}$. The focal length of the mirror used is ((x)/(4)) cm$\left(\frac{x}{4}\right)\text{ cm}$, where magnitude of value of x$x$ is

Related Formula

Magnification formula for thin lens in terms of f$f$ and u$u$:

Core Logic

Since the sizes of the images are equal at two different object positions but the focal length doesn't change, one image must be real and the other virtual. Hence, their magnifications are equal in magnitude but opposite in sign. m₁ = -m₂$m_1 = -m_2$

Step 1: Set Up Magnification Equations

Object distance 1: u₁ = -8 cm$u_1 = -8 \, \mathrm{cm}$ (virtual image case, closer to lens) Object distance 2: u₂ = -24 cm$u_2 = -24 \, \mathrm{cm}$ (real image case)

Using proper sign convention:

Step 2: Solve for Focal Length

Assuming f ≠ 0$f \neq 0$:

f - 24 = -f + 8$f - 24 = -f + 8$ 2f = 32$2f = 32$

Pattern Recognition

For a convex lens, if equal size images are formed at two different object distances u₁$u_1$ and u₂$u_2$, then f = (u₁ + u₂)/(2)$f = \frac{u_1 + u_2}{2}$ if both distances are strictly positive magnitude values.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Related Formula

Core Logic

For minimum deviation, i = e$i = e$ and r = (A)/(2)$r = \frac{A}{2}$. At the exit surface, the light ray exactly meets the condition of the critical angle.

For Total Internal Reflection (TIR) condition matching critical angle: r = θc$r = \theta_c$

Step 1: Solve for Prism Angle

Since r = (1)/(2)$\sin r = \frac{1}{2}$, we have r = 30°$r = 30^{\circ}$. Since r = (A)/(2)$r = \frac{A}{2}$ at minimum deviation:

A = 60°$A = 60^{\circ}$

Pattern Recognition

Critical angle at an interface is always ⁻¹(μᵣₐᵣₑ/μdense)$\sin^{-1}(\mu_{\text{rare}}/\mu_{\text{dense}})$. If μᵣₐᵣₑ$\mu_{\text{rare}}$ is exactly half of μdense$\mu_{\text{dense}}$, the critical angle is exactly 30^°$30^\circ$.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Related Formula

Core Logic

Given parameters: Focal length of objective, f₀ = 2 cm$f_0 = 2 \text{ cm}$ Focal length of eyepiece, fₑ = 5 cm$f_e = 5 \text{ cm}$ Tube length, = 10 cm$\ell = 10 \text{ cm}$ Least distance of distinct vision, D = 25 cm$D = 25 \text{ cm}$

Step 1: Calculate Total Magnification

For normal adjustment (final image at infinity), the magnification formula is:

Since M = 25 = 5²$M = 25 = 5^2$, and the problem states M = (5)^k$M = (5)^k$, we compare to find k=2$k=2$.

Pattern Recognition

For normal adjustment of a compound microscope, M = mₒ × mₑ ≈ (L/fₒ) × (D/fₑ)$M = m_o \times m_e \approx (L/f_o) \times (D/f_e)$.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Related Formula

Core Logic

Given Power P = +5D$P = +5\mathrm{D}$

If the object is between f$f$ and 2f$2f$ (between 20cm$20\text{cm}$ and 40cm$40\text{cm}$), the image must be formed beyond 2f$2f$ (i.e., v > 40cm$v > 40\text{cm}$) and it is magnified.

Step 1: Check P3 Reading

For P₃$P_3$, u = 35cm$u = 35\text{cm}$ (which is between 20cm$20\text{cm}$ and 40cm$40\text{cm}$). Thus, the image distance v$v$ must be greater than 40cm$40\text{cm}$. However, the recorded reading is v = 37cm$v = 37\text{cm}$, which is incorrect.

Pattern Recognition

For a convex lens, if u$u$ lies between f$f$ and 2f$2f$, v$v$ must be greater than 2f$2f$. A simple boundary check on the 2f$2f$ mark (40cm$40\text{cm}$) reveals faulty readings instantly without solving the lens formula.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments

Related Formula

Core Logic

For a real image, magnification m$m$ is negative: m = -3$m = -3$

Given the distance between object and image is 40 cm$40\text{ cm}$: For real images formed by a concave mirror, both u$u$ and v$v$ are on the same side. The distance between them is |v| - |u| = 40 cm$|v| - |u| = 40\text{ cm}$.

Step 1: Determine Object and Image Distances

3|u| - |u| = 40$3|u| - |u| = 40$

Applying sign convention:

Step 2: Compute Focal Length

Pattern Recognition

Real magnified image means concave mirror. The separation is (m-1)|u|$(m-1)|u|$. Therefore |u| = Δ / (m-1)$|u| = \Delta / (m-1)$. Then strictly apply (1)/(f) = (1)/(v) + (1)/(u)$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ with signs.

Chapter Mix

Class 12 Physics: Ray Optics and Optical Instruments