Two identical symmetric double convex lenses of focal length f$f$ are cut into two equal parts L_1, L_2$L_1, L_2$ by AB plane and L_3, L_4$L_3, L_4$ by XY plane as shown in figure respectively. The ratio of focal lengths of lenses L_1$L_1$ and L_3$L_3$ is:

### Related Formula For a prism of angle A$A$: - Deviation: delta = i + e - A$\delta = i + e - A$ - Minimum deviation delta_textmin$\delta_{\text{min}}$ occurs when: i = e quad textand quad r_1 = r_2 = fracA2$$i = e \quad \text{and} \quad r_1 = r_2 = \frac{A}{2}$$ - Under minimum deviation, the ray inside an equilateral/isosceles prism travels symmetrically, making it parallel to the prism base. ### Core Logic Let's check the validity of each statement: - **Statement (A)**: The refracted ray inside the prism becomes parallel to the base at minimum deviation. (True for symmetric prisms) - **Statement (B)**: Larger angle prisms provide smaller minimum deviation. By minimum deviation equation: mu = fracsinleft(fracA+delta_textmin2right)sinleft(fracA2right)$$\mu = \frac{\sin\left(\frac{A+\delta_{\text{min}}}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$ As A$A$ increases, delta_textmin$\delta_{\text{min}}$ generally increases, not decreases. (False) - **Statement (C)**: Angle of incidence i$i$ and angle of emergence e$e$ become equal (i = e$i = e$) during the minimum deviation state. (True) - **Statement (D)**: The delta-i$\delta-i$ curve is asymmetric and parabolic-like; for any deviation delta > delta_textmin$\delta > \delta_{\text{min}}$, there are always exactly two different incident angles (i$i$ and e$e$) that yield the same deviation, except at the minimum deviation point (which has a single unique value). (True) - **Statement (E)**: Angle of refraction r = A/2$r = A/2$, which is half of the prism angle, not double. (False) ### Step 1: Conclusion Since statements A, C, and D are true, the correct option is (1). ### Pattern Recognition Review the classic parabolic shape of the deviation vs. angle of incidence (delta-i$\delta-i$) graph. Notice that any horizontal line above the minimum point intersects twice (representing i$i$ and e$e$ for that deviation). Minimum deviation is the unique local minimum, where i = e$i = e$ and r_1 = r_2 = A/2$r_1 = r_2 = A/2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments: Refraction through Prism

### Related Formula Mirror Formula: frac1v + frac1u = frac1f$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ ### Core Logic For the convex mirror, u = -30text cm$u = -30\text{ cm}$ and f = +30text cm$f = +30\text{ cm}$. frac1v - frac130 = frac130 implies frac1v = frac230 implies v = +15text cm$$\frac{1}{v} - \frac{1}{30} = \frac{1}{30} \implies \frac{1}{v} = \frac{2}{30} \implies v = +15\text{ cm}$$ So, the convex mirror forms a virtual image 15 cm behind its surface. ### Step 1: Align Plane Mirror Image The total distance from the object to the image location is 30 + 15 = 45text cm$30 + 15 = 45\text{ cm}$. For a plane mirror to create an image at this same exact location, it must be placed precisely midway between the object and the image. Distance from object to plane mirror: d = frac452 = 22.5text cm$$d = \frac{45}{2} = 22.5\text{ cm}$$ Therefore, the clearance distance between the convex mirror and the plane mirror surface is: textDistance = 30 - 22.5 = 7.5text cm$$\text{Distance} = 30 - 22.5 = 7.5\text{ cm}$$ ### Pattern Recognition Coinciding images imply identical coordinate endpoints. Calculate the convex position explicitly, find the total path length from the real source object, and slice it in half for the plane mirror location. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula Magnification formula in terms of focal length f$f$ and object position u$u$: m = fracff - u$$m = \frac{f}{f - u}$$ ### Core Logic Case 1: u_1 = -40mathrm~cm$u_1 = -40\mathrm{~cm}$ and m_1 = frac12$m_1 = \frac{1}{2}$ (assuming real inverted image structure for typical convergence calculations): frac12 = fracff - (-40) implies f + 40 = 2f implies f = 40mathrm~cm $$\frac{1}{2} = \frac{f}{f - (-40)} implies f + 40 = 2f implies f = 40\mathrm{~cm} $$ (Taking the magnitude parameter yields focal distance benchmark value). ### Step 1: Calculate New Object Position Case 2: To establish m_2 = frac13$m_2 = \frac{1}{3}$: frac13 = frac4040 - u_2 implies 40 - u_2 = 120 implies u_2 = -80mathrm~cm $$ \frac{1}{3} = \frac{40}{40 - u_2} implies 40 - u_2 = 120 implies u_2 = -80\mathrm{~cm} $$ ### Step 2: Determine Distance Shift Initial location: -40mathrm~cm$-40\mathrm{~cm}$ Final location: -80mathrm~cm$-80\mathrm{~cm}$ textShift = |u_2| - |u_1| = 80 - 40 = 40mathrm~cmtext away from the mirror. $$ \text{Shift} = |u_2| - |u_1| = 80 - 40 = 40\mathrm{~cm}\text{ away from the mirror.} $$ ### Pattern Recognition To reduce the magnification of a real image formed by a concave mirror, the object must always be translated further out away from the focal center point. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula Magnification relation for spherical mirrors: m = -fracvu$$m = -\frac{v}{u}$$ Mirror equation: frac1f = frac1v + frac1u$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$ ### Core Logic Given m = -frac13$m = -\frac{1}{3}$: -fracvu = -frac13 implies u = 3v$$-\frac{v}{u} = -\frac{1}{3} implies u = 3v$$ Since magnification is negative, a real inverted image is formed on the same side as the object in a concave mirror layout configuration.

### Step 1: Formulate Position Distances The distance between the object and image is given as 30mathrm~cm$30\mathrm{~cm}$: |u| - |v| = 30 implies 3v - v = 30 implies 2v = 30 implies v = 15mathrm~cm$$|u| - |v| = 30 implies 3v - v = 30 implies 2v = 30 implies v = 15\mathrm{~cm}$$ This gives u = 3(15) = 45mathrm~cm$u = 3(15) = 45\mathrm{~cm}$. ### Step 2: Solve for Focal Length and x Apply standard sign conventions (u = -45mathrm~cm$u = -45\mathrm{~cm}$, v = -15mathrm~cm$v = -15\mathrm{~cm}$): frac1f = -frac115 - frac145 = frac-3 - 145 = -frac445$$\frac{1}{f} = -\frac{1}{15} - \frac{1}{45} = \frac{-3 - 1}{45} = -\frac{4}{45}$$ |f| = frac454mathrm~cm$$|f| = \frac{45}{4}\mathrm{~cm}$$ Matching this with the prompt pattern form fracx4$\frac{x}{4}$ yields: x = 45$x = 45$ ### Pattern Recognition Real inverted diminished images (|m| < 1$|m| < 1$) mean the image forms closer to the mirror surface than the object, situated between the focal center F$F$ and center of curvature C$C$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments

### Related Formula m = -fracvu$$m = -\frac{v}{u}$$ [cite: 700] frac1v + frac1u = frac1f$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$ [cite: 707] ### Core Logic Given magnification magnitude |m| = frac14$|m| = \frac{1}{4}$[cite: 113]. Assuming a real image formed by a concave mirror: [cite: 701] fracvu = frac14 implies u = 4v$$\frac{v}{u} = \frac{1}{4} \implies u = 4v$$ [cite: 701] The distance between the object and the image is given as 40\ textcm$40\ \text{cm}$ [cite: 114, 116]: u - v = 40 implies 4v - v = 40 implies 3v = 40 implies v = frac403\ textcm$$u - v = 40 \implies 4v - v = 40 \implies 3v = 40 \implies v = \frac{40}{3}\ \text{cm}$$ u = 4 times frac403 = frac1603\ textcm$$u = 4 \times \frac{40}{3} = \frac{160}{3}\ \text{cm}$$ Applying mirror sign conventions (u = -frac1603$u = -\frac{160}{3}$, v = -frac403$v = -\frac{40}{3}$): [cite: 701, 704] frac1f = -frac340 - frac3160 = -frac12 + 3160 = -frac15160$$\frac{1}{f} = -\frac{3}{40} - \frac{3}{160} = -\frac{12 + 3}{160} = -\frac{15}{160}$$ f = -frac16015 approx -10.67\ textcm$$f = -\frac{160}{15} \approx -10.67\ \text{cm}$$ [cite: 712] Rounding to the matching options choice gives 10.7\ textcm$10.7\ \text{cm}$[cite: 118, 712]. ### Pattern Recognition Pay attention to sign conventions in mirror systems. A smaller real image formed by a concave mirror sits between the focus and center of curvature, resulting in u > v$u > v$ configurations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Ray Optics and Optical Instruments