CHAPTER 9: RAY OPTICS AND OPTICAL INSTRUMENTS
SAMPLE QUESTION PAPER (ASSIGNMENT-1)
- There are 33 questions in all. All questions are compulsory.
- Section A: 16 questions (12 MCQs and 4 Assertion-Reasoning) of 1 mark each.
- Section B: 5 short answer questions of 2 marks each.
- Section C: 7 short answer questions of 3 marks each.
- Section D: 3 long answer questions of 5 marks each.
- Section E: 2 case study-based questions of 4 marks each.
SECTION A (16 Marks)
Q.1. The focal length of a plane mirror is:
Explanation: A plane mirror has an infinitely large radius of curvature ($\displaystyle {R = \infty}$), so its focal length $\displaystyle {f = R/2 = \infty}$.
Q.2. A convex mirror is always used as a:
Explanation: Convex mirrors form erect, diminished virtual images and provide a much wider field of view compared to plane mirrors.
Q.3. In the phenomenon of refraction of light, the physical quantity that always remains unchanged is:
Explanation: Frequency is determined solely by the light source and does not alter when light crosses the boundary between different media.
Q.4. The minimum distance at which a normal human eye can see objects clearly without strain is:
Explanation: This distance is known as the Least Distance of Distinct Vision (LDDV), denoted by $\displaystyle {D = 25\text{ cm}}$.
Q.5. Total internal reflection (TIR) takes place when light:
Explanation: Both conditions must be satisfied: (1) light travels from optically denser to rarer medium, and (2) angle of incidence $\displaystyle {i > i_c}$.
Q.6. The blue colour of the clear sky is primarily due to:
Explanation: Rayleigh scattering intensity varies as $\displaystyle {I \propto 1/\lambda^4}$. Shorter blue wavelengths are scattered far more strongly by atmospheric molecules than longer red wavelengths.
Q.7. Dispersion of white light into its constituent colors by a glass prism occurs because of:
Explanation: Refractive index of glass varies with wavelength ($\displaystyle {n_\text{violet} > n_\text{red}}$), causing different colors to refract by different angles.
Q.8. A thin prism of refracting angle $\displaystyle {A = 6^\circ}$ is made of glass of refractive index $\displaystyle {n = 1.60}$. The angle of minimum deviation $\displaystyle {\delta_{\text{min}}}$ is:
Explanation: For a thin prism, deviation $\displaystyle {\delta \approx (n - 1) A = (1.60 - 1) \times 6^\circ = 0.60 \times 6^\circ = 3.6^\circ}$.
Q.9. A simple microscope uses a convex lens of focal length $\displaystyle {f = 5\text{ cm}}$. If the final image is formed at LDDV ($\displaystyle {D = 25\text{ cm}}$), its magnifying power $\displaystyle {M}$ is:
Explanation: Magnifying power at near point: $\displaystyle {M = 1 + {\frac{{D}}{{f}}} = 1 + {\frac{{25}}{{5}}} = 1 + 5 = 6}$.
Q.10. A compound microscope has objective focal length $\displaystyle {f_o = 1\text{ cm}}$, eyepiece focal length $\displaystyle {f_e = 5\text{ cm}}$, and tube length $\displaystyle {L = 20\text{ cm}}$. For final image at $\displaystyle {D = 25\text{ cm}}$, magnification $\displaystyle {M}$ is:
Explanation: $\displaystyle {M \approx {\left({\frac{{L}}{{f_o}}}\right)} {\left({\frac{{D}}{{f_e}}}\right)} = {\left({\frac{{20}}{{1}}}\right)} {\left({\frac{{25}}{{5}}}\right)} = 20 \times 5 = 100}$.
Q.11. An astronomical telescope in normal adjustment has objective focal length $\displaystyle {f_o = 100\text{ cm}}$ and eyepiece focal length $\displaystyle {f_e = 5\text{ cm}}$. Its magnifying power $\displaystyle {m}$ is:
Explanation: For normal adjustment (final image at infinity), $\displaystyle {m = {\frac{{f_o}}{{f_e}}} = {\frac{{100}}{{5}}} = 20}$.
Q.12. Light travels from air ($\displaystyle {n_1 = 1}$) into glass ($\displaystyle {n_2 = 1.5}$) through a convex spherical surface of radius $\displaystyle {R = +20\text{ cm}}$. An object is placed in air at $\displaystyle {u = -30\text{ cm}}$. The image distance $\displaystyle {v}$ is:
Explanation: Refraction formula at spherical surface: $\displaystyle {{\frac{{n_2}}{{v}}} - {\frac{{n_1}}{{u}}} = {\frac{{n_2 - n_1}}{{R}}} \implies {\frac{{1.5}}{{v}}} - {\frac{{1}}{{-30}}} = {\frac{{1.5 - 1.0}}{{20}}}}$.
$\displaystyle {{\frac{{1.5}}{{v}}} + {\frac{{1}}{{30}}} = {\frac{{1}}{{40}}} \implies {\frac{{1.5}}{{v}}} = {\frac{{1}}{{40}}} - {\frac{{1}}{{30}}} = {\frac{{3 - 4}}{{120}}} = -{\frac{{1}}{{120}}} \implies v = 1.5 \times (-120) = -180\text{ cm}}$.
Directions for Q.13 to Q.16:
(A) Both Assertion and Reason are true and Reason is correct explanation of A.
(B) Both Assertion and Reason are true but Reason is NOT correct explanation of A.
(C) Assertion is true but Reason is false.
(D) Both Assertion and Reason are false.
Q.13. Assertion (A): Concave mirrors are used by dentists to examine teeth.
Reason (R): Concave mirrors form erect and magnified virtual images when the object is placed close to the mirror (between focus and pole).
Q.14. Assertion (A): In a compound microscope, the total magnification depends on the focal lengths of both objective and eyepiece.
Reason (R): Total magnification is equal to the product of magnification produced by objective and eyepiece ($\displaystyle {m = m_o \times m_e}$).
Q.15. Assertion (A): In an astronomical telescope, the objective lens has a much larger focal length than the eyepiece.
Reason (R): The objective is used to collect light and form a real image, which is then magnified by the eyepiece.
Q.16. Assertion (A): A convex mirror always produces a diminished virtual image of a real object.
Reason (R): The focal length of a convex mirror is positive by sign convention.
Explanation: Both statements are true, but the positive focal length sign convention is not the physical reason why a convex mirror forms diminished images.
SECTION B (10 Marks)
Q.17. Why does a glass prism produce a spectrum of white light, whereas a parallel-sided glass slab does not?
Why are reflecting telescopes preferred over refracting telescopes? Justify with at least three points.
In a glass prism, refracting faces are non-parallel, so emergent rays for different colors bend by different total angles of deviation ($\displaystyle {\delta = (n-1)A}$). In a parallel glass slab, dispersion occurs at the first face but is reversed at the parallel second face, so emergent rays of all colors emerge parallel to each other with only lateral displacement.
OR Part (Reflecting Telescopes Advantages):
1. Free from chromatic aberration since reflection obeys laws of reflection independent of wavelength.
2. Spherical aberration is eliminated by using parabolic primary mirrors.
3. Parabolic mirrors can be supported easily over their entire back surface and have high light-gathering power.
Q.18. A solid glass sphere of radius $\displaystyle {R = 6.0\text{ cm}}$ ($\displaystyle {n = 1.5}$) has a small air bubble trapped at distance $\displaystyle {3.0\text{ cm}}$ from its center $\displaystyle {C}$. Find the apparent position of the bubble when viewed through the surface from an outside point $\displaystyle {E}$ in air.
Light travels from glass ($\displaystyle {n_1 = 1.5}$) to air ($\displaystyle {n_2 = 1.0}$). Surface is concave towards denser medium ($\displaystyle {R = -6.0\text{ cm}}$).
Formula: $\displaystyle {{\frac{{n_2}}{{v}}} - {\frac{{n_1}}{{u}}} = {\frac{{n_2 - n_1}}{{R}}}} \implies \displaystyle {{\frac{{1.0}}{{v}}} - {\frac{{1.5}}{{-3.0}}} = {\frac{{1.0 - 1.5}}{{-6.0}}}}$.
$\displaystyle {{\frac{{1}}{{v}}} + 0.5 = {\frac{{0.5}}{{6.0}}} = {\frac{{1}}{{12}}} \implies {\frac{{1}}{{v}}} = {\frac{{1}}{{12}}} - {\frac{{1}}{{2}}} = -{\frac{{5}}{{12}}} \implies v = -2.4\text{ cm}}$.
The bubble appears at a depth of $\displaystyle {2.4\text{ cm}}$ inside the surface.
Q.19. A double convex lens is made of glass of refractive index $\displaystyle {1.55}$, with both faces having equal radius of curvature $\displaystyle {R}$. Find $\displaystyle {R}$ if focal length is $\displaystyle {20\text{ cm}}$.
Given three lenses: $\displaystyle {L_1 (3\text{ D}, 8\text{ cm aperture})}$, $\displaystyle {L_2 (6\text{ D}, 1\text{ cm aperture})}$, $\displaystyle {L_3 (10\text{ D}, 1\text{ cm aperture})}$. Which two lenses will you choose as objective and eyepiece to construct an astronomical telescope? Justify.
Lens Maker's Formula: $\displaystyle {{\frac{{1}}{{f}}} = (n - 1) {\left( {\frac{{1}}{{R_1}}} - {\frac{{1}}{{R_2}}} \right)}}$. Here $\displaystyle {R_1 = +R, R_2 = -R}$.
$\displaystyle {{\frac{{1}}{{20}}} = (1.55 - 1) {\left( {\frac{{1}}{{R}} + {\frac{{1}}{{R}}} \right)} = 0.55 \times {\frac{{2}}{{R}}} = {\frac{{1.10}}{{R}}} \implies R = 20 \times 1.10 = 22\text{ cm}}$.
OR Part:
Objective: Lens $\displaystyle {L_1}$ ($\displaystyle {P = 3\text{ D} \implies f_o = 33.3\text{ cm}}$, large aperture $\displaystyle {8\text{ cm}}$ to collect maximum light).
Eyepiece: Lens $\displaystyle {L_3}$ ($\displaystyle {P = 10\text{ D} \implies f_e = 10\text{ cm}}$, small focal length and small aperture $\displaystyle {1\text{ cm}}$).
Q.20. State the condition under which large magnification can be achieved in an astronomical telescope.
1. Focal length of the objective lens $\displaystyle {f_o}$ must be as large as possible.
2. Focal length of the eyepiece lens $\displaystyle {f_e}$ must be as small as possible.
Q.21. The focal length of an equi-convex lens is equal to the radius of curvature of either face. What is the refractive index of the lens material?
By Lens Maker's Formula: $\displaystyle {{\frac{{1}}{{f}}} = (n - 1) {\left( {\frac{{1}}{{R_1}}} - {\frac{{1}}{{R_2}}} \right)} \implies {\frac{{1}}{{R}}} = (n - 1) {\left( {\frac{{2}}{{R}}} \right)}}$.
$\displaystyle {1 = 2 (n - 1) \implies n - 1 = 0.5 \implies n = 1.5}$.
SECTION C (21 Marks)
Q.22. A triangular prism of refracting angle $\displaystyle {A = 60^\circ}$ is made of transparent material of refractive index $\displaystyle {n = 2/\sqrt{3}}$. A ray of light is incident normally on face KL as shown below. Trace the path of ray, calculate angle of emergence and angle of deviation.
At face KM, angle of incidence $\displaystyle {r_2 = A - r_1 = 60^\circ - 0^\circ = 60^\circ}$.
Critical angle for prism: $\displaystyle {\sin i_c = 1/n = \sqrt{3}/2 \implies i_c = 60^\circ}$.
Since $\displaystyle {r_2 = i_c = 60^\circ}$, the ray emerges grazing the surface KM with angle of emergence $\displaystyle {e = 90^\circ}$.
Total Angle of Deviation: $\displaystyle {\delta = i + e - A = 0^\circ + 90^\circ - 60^\circ = 30^\circ}$.
Q.23. A plano-convex and a plano-concave lens each of radius of curvature $\displaystyle {R = 15\text{ cm}}$ and refractive index $\displaystyle {n = 1.5}$ are placed together. Find the final image position for parallel incident rays.
Complete the path of parallel light rays emerging from a convex lens of material $\displaystyle {n_1}$ placed in medium $\displaystyle {n_2}$ for: (i) $\displaystyle {n_1 > n_2}$, (ii) $\displaystyle {n_1 = n_2}$, (iii) $\displaystyle {n_1 < n_2}$.
For plano-convex lens: $\displaystyle {1/f_1 = (1.5 - 1)(1/15 - 1/\infty) = +1/30 \implies f_1 = +30\text{ cm}}$.
For plano-concave lens: $\displaystyle {1/f_2 = (1.5 - 1)(1/\infty - 1/15) = -1/30 \implies f_2 = -30\text{ cm}}$.
Combined focal length: $\displaystyle {1/F = 1/f_1 + 1/f_2 = 1/30 - 1/30 = 0 \implies F = \infty}$. Parallel rays emerge parallel ($\displaystyle {v = \infty}$).
OR Part:
(i) $\displaystyle {n_1 > n_2}$: Lens behaves as a normal converging lens.
(ii) $\displaystyle {n_1 = n_2}$: Lens behaves as a simple glass plate; light passes unbent without refraction.
(iii) $\displaystyle {n_1 < n_2}$: Lens reverses its nature and behaves as a diverging lens.
Q.24. An object is placed in front of a concave mirror such that a virtual image is formed. Draw a neat labeled ray diagram and derive the mirror formula $\displaystyle {1/f = 1/v + 1/u}$ and linear magnification $\displaystyle {m = -v/u}$.
Applying sign convention: $\displaystyle {AB = +h, A'B' = +h', PB = -u, PB' = +v}$.
Magnification: $\displaystyle {m = {\frac{{h'}}{{h}}} = {\frac{{+v}}{{-u}}} \implies m = -{\frac{{v}}{{u}}}}$.
Using similar triangles with focus $\displaystyle {F}$, substitution yields mirror formula: $\displaystyle {{\frac{{1}}{{f}}} = {\frac{{1}}{{v}}} + {\frac{{1}}{{u}}}}$.
Q.25. Derive Lens Maker's Formula $\displaystyle {1/f = (n - 1)(1/R_1 - 1/R_2)}$ by drawing a neat ray diagram for refraction at two spherical surfaces of a thin convex lens, and hence obtain thin lens formula.
Refraction at 2nd surface (virtual object $\displaystyle {I'}$): $\displaystyle {{\frac{{n_1}}{{v}}} - {\frac{{n_2}}{{v'}}} = {\frac{{n_1 - n_2}}{{R_2}}} = -{\frac{{n_2 - n_1}}{{R_2}}}}$.
Adding equations: $\displaystyle {n_1 {\left( {\frac{{1}}{{v}}} - {\frac{{1}}{{u}}} \right)} = (n_2 - n_1) {\left( {\frac{{1}}{{R_1}}} - {\frac{{1}}{{R_2}}} \right)} \implies {\frac{{1}}{{v}}} - {\frac{{1}}{{u}}} = {\left( {\frac{{n_2}}{{n_1}}} - 1 \right)} {\left( {\frac{{1}}{{R_1}}} - {\frac{{1}}{{R_2}}} \right)}}$.
For $\displaystyle {u = \infty, v = f}$: $\displaystyle {{\frac{{1}}{{f}}} = (n - 1) {\left( {\frac{{1}}{{R_1}}} - {\frac{{1}}{{R_2}}} \right)}}$. Comparing gives thin lens formula $\displaystyle {{\frac{{1}}{{v}}} - {\frac{{1}}{{u}}} = {\frac{{1}}{{f}}}}$.
Q.26. Two thin lenses $\displaystyle {L_1}$ (convex, $\displaystyle {f_1 = +24\text{ cm}}$) and $\displaystyle {L_2}$ (concave, $\displaystyle {f_2 = -18\text{ cm}}$) are placed coaxially $\displaystyle {45\text{ cm}}$ apart. A $\displaystyle {1.0\text{ cm}}$ tall object is placed $\displaystyle {36\text{ cm}}$ in front of $\displaystyle {L_1}$. Find location and height of final image.
$\displaystyle {{\frac{{1}}{{v_1}}} = {\frac{{1}}{{f_1}}} + {\frac{{1}}{{u_1}}} = {\frac{{1}}{{24}}} - {\frac{{1}}{{36}}} = {\frac{{3 - 2}}{{72}}} = {\frac{{1}}{{72}}} \implies v_1 = +72\text{ cm}}$.
This real image forms $\displaystyle {72\text{ cm}}$ right of $\displaystyle {L_1}$, which acts as virtual object for $\displaystyle {L_2}$ at $\displaystyle {u_2 = +(72 - 45) = +27\text{ cm}}$.
For lens $\displaystyle {L_2}$: $\displaystyle {u_2 = +27\text{ cm}, f_2 = -18\text{ cm}}$.
$\displaystyle {{\frac{{1}}{{v_2}}} = {\frac{{1}}{{f_2}}} + {\frac{{1}}{{u_2}}} = -{\frac{{1}}{{18}}} + {\frac{{1}}{{27}}} = {\frac{{-3 + 2}}{{54}}} = -{\frac{{1}}{{54}}} \implies v_2 = -54\text{ cm}}$.
Final image is formed $\displaystyle {54\text{ cm}}$ to the left of lens $\displaystyle {L_2}$.
Magnification $\displaystyle {m = m_1 \times m_2 = {\left({\frac{{v_1}}{{u_1}}}\right)} {\left({\frac{{v_2}}{{u_2}}}\right)} = {\left({\frac{{72}}{{-36}}}\right)} {\left({\frac{{-54}}{{27}}}\right)} = (-2) \times (-2) = +4}$. Height $\displaystyle {h' = +4\text{ cm}}$.
Q.27. You are given three lenses of powers $\displaystyle {5\text{ D}}$, $\displaystyle {4\text{ D}}$, and $\displaystyle {10\text{ D}}$ to design an astronomical telescope.
(a) Which lenses should be used as objective and eyepiece? Justify.
(b) Why is the aperture of the objective preferred to be large?
(a) Objective: Lens with $\displaystyle {P = 4\text{ D}}$ ($\displaystyle {f_o = 1/4 = 25\text{ cm}}$, largest focal length).
Eyepiece: Lens with $\displaystyle {P = 10\text{ D}}$ ($\displaystyle {f_e = 1/10 = 10\text{ cm}}$, smallest focal length).
This gives highest magnifying power $\displaystyle {m = f_o / f_e = P_e / P_o = 10/4 = 2.5}$.
(b) A large aperture objective collects more light from distant objects to form brighter images and increases resolving power ($\displaystyle {R.P. \propto D/\lambda}$).
Q.28. A point $\displaystyle {O}$ marked on the surface of a glass sphere of diameter $\displaystyle {20\text{ cm}}$ ($\displaystyle {n = 1.5}$) is viewed through glass from the position directly opposite to $\displaystyle {O}$. Find the image position and draw ray diagram.
Three rays (1, 2, 3) of different colors fall normally on face AB of an isosceles right-angled prism ABC ($\displaystyle {45^\circ-90^\circ-45^\circ}$). Refractive indices are $\displaystyle {1.39, 1.47, 1.52}$ respectively. Which rays undergo TIR at face AC?
Light travels from glass ($\displaystyle {n_1 = 1.5}$) to air ($\displaystyle {n_2 = 1.0}$) across concave surface ($\displaystyle {R = -10\text{ cm}}$, object at $\displaystyle {u = -20\text{ cm}}$).
Formula: $\displaystyle {{\frac{{n_2}}{{v}}} - {\frac{{n_1}}{{u}}} = {\frac{{n_2 - n_1}}{{R}}} \implies {\frac{{1.0}}{{v}}} - {\frac{{1.5}}{{-20}}} = {\frac{{1.0 - 1.5}}{{-10}}}}$.
$\displaystyle {{\frac{{1}}{{v}}} + {\frac{{3}}{{40}}} = {\frac{{1}}{{20}}} \implies {\frac{{1}}{{v}}} = {\frac{{1}}{{20}}} - {\frac{{3}}{{40}}} = -{\frac{{1}}{{40}}} \implies v = -40\text{ cm}}$.
Image is virtual, formed $\displaystyle {40\text{ cm}}$ behind the refracting surface.
OR Part: At face AC, angle of incidence $\displaystyle {i = 45^\circ}$. For TIR, $\displaystyle {i > i_c \implies \sin 45^\circ > 1/n \implies 1/\sqrt{2} > 1/n \implies n > \sqrt{2} \approx 1.414}$.
Rays 2 ($\displaystyle {n = 1.47}$) and 3 ($\displaystyle {n = 1.52}$) have $\displaystyle {n > 1.414}$ and undergo Total Internal Reflection. Ray 1 ($\displaystyle {n = 1.39}$) refracts into air.
SECTION D: LONG ANSWER QUESTIONS (15 Marks)
Q.29. (i) Draw a labeled ray diagram for image formation by a compound microscope. Write expressions for magnifying power when final image is formed at near point and at infinity.
(ii) Why must both objective and eyepiece have short focal lengths?
Total magnification of a compound microscope is 20. Eyepiece magnification is 5. Microscope is focused on an object with tube length $\displaystyle {L = 14\text{ cm}}$ and $\displaystyle {D = 20\text{ cm}}$. Calculate focal length of objective $\displaystyle {f_o}$ and eyepiece $\displaystyle {f_e}$.
(i) Magnification at near point: $\displaystyle {M = -{\frac{{v_o}}{{u_o}}} {\left( 1 + {\frac{{D}}{{f_e}}} \right)} \approx -{\frac{{L}}{{f_o}}} {\left( 1 + {\frac{{D}}{{f_e}}} \right)}}$. At infinity: $\displaystyle {M = -{\frac{{L D}}{{f_o f_e}}}}$.
(ii) Both focal lengths must be short to maximize total angular magnification ($\displaystyle {M \propto 1/(f_o f_e)}$).
OR Part: Given $\displaystyle {M = -20, m_e = 5, v_e = -D = -20\text{ cm}}$.
For eyepiece: $\displaystyle {m_e = v_e / u_e \implies 5 = -20 / u_e \implies u_e = -4\text{ cm}}$.
Lens formula for eyepiece: $\displaystyle {1/v_e - 1/u_e = 1/f_e \implies -1/20 + 1/4 = 1/f_e \implies 4/20 = 1/f_e \implies f_e = 5\text{ cm}}$.
Total magnification: $\displaystyle {M = m_o \times m_e \implies -20 = m_o \times 5 \implies m_o = -4}$.
Tube length $\displaystyle {L = v_o + |u_e| \implies 14 = v_o + 4 \implies v_o = 10\text{ cm}}$.
$\displaystyle {m_o = 1 - v_o/f_o \implies -4 = 1 - 10/f_o \implies 10/f_o = 5 \implies f_o = 2\text{ cm}}$.
Q.30. Draw ray diagram showing image formation of a distant object by a refracting astronomical telescope. Write magnifying power and tube length expressions for: (i) LDDV, (ii) normal adjustment. Mention two factors to increase magnifying power.
(i) State main considerations when choosing objective of an astronomical telescope.
(ii) Draw ray diagram of a reflecting type (Cassegrain) telescope and state its magnifying power.
(i) Final image at LDDV ($\displaystyle {D}$): $\displaystyle {m = -{\frac{{f_o}}{{f_e}}} {\left( 1 + {\frac{{f_e}}{{D}}} \right)}}$, Tube length $\displaystyle {L = f_o + u_e}$.
(ii) Normal adjustment (at infinity): $\displaystyle {m = -{\frac{{f_o}}{{f_e}}}}$, Tube length $\displaystyle {L = f_o + f_e}$.
Factors to increase $m$: Increase objective focal length $\displaystyle {f_o}$; decrease eyepiece focal length $\displaystyle {f_e}$.
OR Part (Cassegrain Reflecting Telescope): Magnifying power $\displaystyle {m \approx f_o / f_e}$. Primary concave mirror collects light, secondary convex mirror reflects rays through central hole to eyepiece.
Q.31. (i) Derive refractive index formula $\displaystyle {n = \frac{\sin[(A + \delta_m)/2]}{\sin(A/2)}}$ for a glass prism and deduce thin prism formula $\displaystyle {\delta_m = (n-1)A}$.
(ii) Draw graph showing variation of angle of deviation $\displaystyle {\delta}$ with angle of incidence $\displaystyle {i}$.
A ray of light passing from air through an equilateral glass prism undergoes minimum deviation when angle of incidence is $\displaystyle {3/4\text{th}}$ of angle of prism. Calculate speed of light in prism.
At minimum deviation: $\displaystyle {i = e}$ and $\displaystyle {r_1 = r_2 = r = A/2}$. Deviation $\displaystyle {\delta_m = 2i - A \implies i = (A + \delta_m)/2}$.
By Snell's Law: $\displaystyle {n = {\frac{{\sin i}}{{\sin r}}} = {\frac{{\sin[(A + \delta_m)/2]}}{{\sin(A/2)}}}$. For thin prism: $\displaystyle {n \approx {\frac{{(A + \delta_m)/2}}{{A/2}}} \implies \delta_m = (n - 1)A}$.
OR Part: Equilateral prism $\displaystyle {A = 60^\circ}$. Given $\displaystyle {i = {\frac{{3}}{{4}}} A = {\frac{{3}}{{4}}} \times 60^\circ = 45^\circ}$.
At minimum deviation: $\displaystyle {r = A/2 = 30^\circ}$.
$\displaystyle {n = {\frac{{\sin 45^\circ}}{{\sin 30^\circ}}} = {\frac{{1/\sqrt{2}}}{{1/2}}} = \sqrt{2} \approx 1.414}$.
Speed of light in prism: $\displaystyle {v = {\frac{{c}}{{n}}} = {\frac{{3 \times 10^8}}{{\sqrt{2}}}} \approx 2.12 \times 10^8\text{ m/s}}$.
SECTION E: CASE STUDY QUESTIONS (8 Marks)
Q.32. Case Study 1: Optical Fibers and Total Internal Reflection.
Optical fibers carry data as light pulses along glass/plastic strands using total internal reflection. A core of higher refractive index is surrounded by cladding of lower refractive index.
(i) When light enters from denser to rarer medium, it bends: (a) Towards normal (b) Away from normal
(ii) Critical angle of water ($\displaystyle {n=1.33}$) in air is approx: (a) $\displaystyle {48.8^\circ}$ (b) $\displaystyle {42^\circ}$ (c) $\displaystyle {50^\circ}$
(iii) Outer concentric layer in optical fiber is called: (a) Cladding (b) Core (c) Mantle
(iv) Critical angle for glass-air ($\displaystyle {n=1.5}$) is: (a) $\displaystyle {42^\circ}$ (b) $\displaystyle {50^\circ}$ (c) $\displaystyle {30^\circ}$
(v) Critical angle for glass-air interface is minimum for: (a) Red (b) Green (c) Violet
Q.33. Case Study 2: Refraction through Lenses & Split Lens.
A convex lens converges parallel rays to a real focus. A point object $\displaystyle {O}$ is placed $\displaystyle {30\text{ cm}}$ from a convex lens ($\displaystyle {f = 20\text{ cm}}$) cut horizontally into two halves displaced vertically by $\displaystyle {2 \times 0.05\text{ cm}}$.
(i) Image position from lens: (a) $\displaystyle {30\text{ cm}}$ right (b) $\displaystyle {60\text{ cm}}$ right (c) $\displaystyle {40\text{ cm}}$ left
(ii) Two thin lenses in contact have combined focal length $\displaystyle {80\text{ cm}}$. If $\displaystyle {f_1 = 20\text{ cm}}$, $\displaystyle {f_2}$ is: (a) $\displaystyle {-26.7\text{ cm}}$ (b) $\displaystyle {60\text{ cm}}$
(iii) Spherical air bubble inside glass behaves like: (a) Converging lens (b) Diverging lens
(iv) Lens used in magnifying glass is: (a) Concave lens (b) Convex lens
(v) Magnification by convex lens is positive when object is: (a) At F (b) Between F and optical center
(i) (b) $\displaystyle {60\text{ cm}}$ right ($\displaystyle {1/v - 1/(-30) = 1/20 \implies 1/v = 1/20 - 1/30 = 1/60 \implies v = +60\text{ cm}}$). Both halves form images in the same vertical plane.
(ii) (a) $\displaystyle {-26.7\text{ cm}}$ ($\displaystyle {1/80 = 1/20 + 1/f_2 \implies 1/f_2 = 1/80 - 1/20 = -3/80 \implies f_2 = -26.7\text{ cm}}$).
(iii) (b) Diverging lens ($\displaystyle {n_\text{air} < n_\text{glass}}$).
(iv) (b) Convex lens.
(v) (d) Between F and optical center (forms erect virtual image).
ASSIGNMENT – 2 (MCQ PRACTICE)
1. The speed of light in a medium is $\displaystyle {2 \times 10^8\text{ m/s}}$. The refractive index of the medium is:
2. If angle of incidence is $\displaystyle {45^\circ}$ and angle of refraction is $\displaystyle {30^\circ}$, refractive index of medium is:
3. A convex lens of focal length $\displaystyle {20\text{ cm}}$ is placed in air. Its power is:
4. Which of the following lenses is always diverging regardless of object position?
5. A ray of light falls normally on a plane mirror. The angle of reflection is:
ASSIGNMENT – 3 (ASSERTION & REASON)
1. Assertion (A): In total internal reflection, light must travel from a denser medium to a rarer medium.
Reason (R): Critical angle exists only when light travels from rarer to denser medium.
Explanation: Critical angle exists only when light travels from denser to rarer medium ($\displaystyle {\sin i_c = n_2/n_1 < 1}$).
2. Assertion (A): For a thin prism, deviation produced is directly proportional to the prism angle.
Reason (R): In thin prism approximation ($\displaystyle {\sin\theta \approx \theta}$), deviation is $\displaystyle {\delta = (n-1)A}$.
ASSIGNMENT – 4 (CASE STUDY QUESTIONS)
Case Study: Compound Microscope Magnification
A compound microscope consists of an objective ($\displaystyle {f_o = 2.0\text{ cm}}$) and eyepiece ($\displaystyle {f_e = 6.25\text{ cm}}$) separated by $\displaystyle {15\text{ cm}}$. Final image is formed at $\displaystyle {D = 25\text{ cm}}$.
1. Object distance for eyepiece ($\displaystyle {u_e}$): (a) $\displaystyle {3.45\text{ cm}}$ (b) $\displaystyle {-5.0\text{ cm}}$
2. Object distance for objective ($\displaystyle {u_o}$): (a) $\displaystyle {-4.5\text{ cm}}$ (b) $\displaystyle {-2.5\text{ cm}}$
3. Nature of intermediate image formed by objective: (a) Real, inverted, magnified (b) Virtual, erect, magnified
1: (b) $\displaystyle {-5.0\text{ cm}}$ ($\displaystyle {1/(-25) - 1/u_e = 1/6.25 \implies u_e = -5\text{ cm}}$).
2: (b) $\displaystyle {-2.5\text{ cm}}$ ($\displaystyle {v_o = L - |u_e| = 15 - 5 = 10\text{ cm} \implies 1/10 - 1/u_o = 1/2 \implies u_o = -2.5\text{ cm}}$).
3: (a) Real, inverted, and magnified.
ASSIGNMENT – 5 (FORMULAE & CONCEPTUAL QUESTIONS)
CORE FORMULAE SUMMARY
Mirror: $\displaystyle {{\frac{{1}}{{f}}} = {\frac{{1}}{{v}}} + {\frac{{1}}{{u}}}}$, Magnification: $\displaystyle {m = -{\frac{{v}}{{u}}}}$
Thin Lens: $\displaystyle {{\frac{{1}}{{f}}} = {\frac{{1}}{{v}}} - {\frac{{1}}{{u}}}}$, Magnification: $\displaystyle {m = +{\frac{{v}}{{u}}}}$
$\displaystyle {{\frac{{1}}{{f}}} = (n - 1) {\left( {\frac{{1}}{{R_1}}} - {\frac{{1}}{{R_2}}} \right)}}$ | Power: $\displaystyle {P = {\frac{{1}}{{f\text{ (m)}}}}}$ (Dioptres, D)
Spherical Surface: $\displaystyle {{\frac{{n_2}}{{v}}} - {\frac{{n_1}}{{u}}} = {\frac{{n_2 - n_1}}{{R}}}}$ | Prism Refractive Index: $\displaystyle {n = {\frac{{\sin[(A + \delta_m)/2]}}{{\sin(A/2)}}}}$
Compound Microscope: $\displaystyle {M = -{\left({\frac{{L}}{{f_o}}}\right)} {\left( 1 + {\frac{{D}}{{f_e}}} \right)}}$ | Telescope (Normal Adjustment): $\displaystyle {m = -{\frac{{f_o}}{{f_e}}}}$, Tube Length: $\displaystyle {L = f_o + f_e}$
CONCEPTUAL SHORT QUESTIONS
1. A convex lens of focal length $\displaystyle {20\text{ cm}}$ in air is immersed in water ($\displaystyle {n_w = 4/3}$). How does its focal length change? ($\displaystyle {n_g = 1.5}$)
In water: $\displaystyle {1/f_w = (1.5/(4/3) - 1)(2/R) = (9/8 - 1)(2/R) = (1/8)(2/R) = 1/(4R) \implies f_w = 4 R = 4 \times 20 = 80\text{ cm}}$.
Focal length increases by 4 times to $\displaystyle {80\text{ cm}}$.
2. Under what condition does a convex lens act as a diverging lens?
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