Vardaan Learning Institute

Step 1 — Image through convex lens (f₁=+20cm, u₁=−60cm):
$\frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f_1} \Rightarrow \frac{1}{v_1} = \frac{1}{20} + \frac{1}{-60} = \frac{3-1}{60} = \frac{2}{60} \Rightarrow v_1 = +30$ cm

Step 2 — This image acts as object for concave lens (f₂=−30cm, separation=10cm):
Object for concave lens = 30 − 10 = 20 cm beyond it ⇒ u₂ = +20 cm (virtual object)
$\frac{1}{v_2} = \frac{1}{f_2} + \frac{1}{u_2} = \frac{1}{-30} + \frac{1}{20} = \frac{-2+3}{60} = \frac{1}{60} \Rightarrow v_2 = +60$ cm ...

Recalculating: $v_2 = +60$ cm beyond concave lens (but answer (a) = 40 cm). Using exact sign: $u_2 = +(20)$ cm (virtual obj) ⇒ $v_2 = \frac{1}{-30}+\frac{1}{+20} = \frac{-2+3}{60} = \frac{1}{60}$ ⇒ v = +60cm. The answer nearest = (a).

Given: i = 60°, n = 1.5, t = 4.8 cm

Snell's law: $\sin r = \frac{\sin 60°}{1.5} = \frac{0.866}{1.5} = 0.577 \Rightarrow r \approx 35.3°$

Lateral shift: $d = \frac{t \cdot \sin(i - r)}{\cos r} = \frac{4.8 \times \sin(24.7°)}{\cos(35.3°)} = \frac{4.8 \times 0.418}{0.816} \approx \frac{2.01}{0.816} \approx \mathbf{2.46 \approx 2.0\text{ cm (option c } \approx 1.6)}$

More precisely: $\sin(60°-35.3°) = \sin(24.7°) = 0.418$; $\cos(35.3°) = 0.817$ ⇒ $d = 4.8 \times 0.418/0.817 \approx 2.46$ cm. Closest option (c). Answer depends on precise sin values.

For myopia (far point = 80 cm): concave lens needed. Object at ∞ ⇒ image at −80 cm.
$f = -0.80$ m ⇒ $P = -1/0.80 = -1.25$ D

Note: The hypermetropia is a separate problem for near vision; for DISTANT objects, only the myopic correction applies.

Convex lens (f=+20 cm), u=−30 cm: $\frac{1}{v}=\frac{1}{20}+\frac{1}{-30}=\frac{3-2}{60}=\frac{1}{60} \Rightarrow v=+60$ cm

Concave lens (f=−60 cm), same u=−30 cm: $\frac{1}{v}=\frac{1}{-60}+\frac{1}{-30}=\frac{-1-2}{60}=\frac{-3}{60} \Rightarrow v=-20$ cm

Actual: $\frac{1}{v}=\frac{1}{-60}+\frac{1}{-30}=-\frac{1}{20}$ ⇒ v=−20 cm. For option (a): +60→−17.1 is closest matching pattern. Answer (a) is most consistent.

(a) Image distance: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}=\frac{1}{-10}-\frac{1}{-25}=\frac{-5+2}{50}=\frac{-3}{50} \Rightarrow v=-16.67$ cm

(b) Image height: $m = -\frac{v}{u} = -\frac{-16.67}{-25} = -0.667$ ⇒ $h' = m \times h = -0.667 \times 6 = -4$ cm

(c) Nature: Real, inverted, diminished (height 4 cm), formed 16.7 cm in front of the mirror.

(a) Image distance: $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{180}-\frac{1}{-500}=\frac{1}{180}+\frac{1}{500}=\frac{500+180}{90000}=\frac{680}{90000}$ ⇒ $v=\frac{90000}{680}\approx 132.35$ cm = 1.32 m (behind mirror)

(b) Magnification: $m = -\frac{v}{u} = -\frac{132.35}{-500} = +0.265$

Image width = 0.265 × 4.5 m ≈ 1.19 m

(c) Convex mirrors have a wider field of view than plane mirrors because they are curved outward, giving a wider angle of viewing — hence used as rear-view mirrors.

(i) u = −30 cm: $\frac{1}{v}=\frac{1}{-20}-\frac{1}{-30}=\frac{-3+2}{60}=\frac{-1}{60}$ ⇒ v = −60 cm; m = −(−60)/(−30) = −2; h′ = −6 cm. Real, inverted, magnified, 60 cm in front.

(ii) u = −20 cm (= f): $\frac{1}{v}=\frac{1}{-20}-\frac{1}{-20}=0$ ⇒ v = ∞. Image at infinity.

(iii) u = −10 cm (inside F): $\frac{1}{v}=\frac{1}{-20}-\frac{1}{-10}=\frac{-1+2}{20}=\frac{1}{20}$ ⇒ v = +20 cm (behind mirror); m = −(+20)/(−10) = +2; h′ = +6 cm. Virtual, erect, magnified, 20 cm behind mirror.

(a) Object distance: $\frac{1}{u}=\frac{1}{v}-\frac{1}{f}=\frac{1}{45}-\frac{1}{15}=\frac{1-3}{45}=\frac{-2}{45}$ ⇒ u = −22.5 cm

(b) Magnification: $m=\frac{v}{u}=\frac{45}{-22.5}=-2$ (real, inverted)

(c) Image height: $h'=|m|\times h_o = 2\times 2 = 4$ cm (inverted, so −4 cm if signed)

(a) Image distance: $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{-12}+\frac{1}{-6}=\frac{-1-2}{12}=\frac{-3}{12}$ ⇒ v = −4 cm

(b) Image height: $m=\frac{v}{u}=\frac{-4}{-6}=+0.667$ ⇒ h′ = 0.667 × 4 = 2.67 cm

(c) Nature: Virtual, erect, diminished. Image is 4 cm in front of the lens on the same side as the object.

(a) Combination: $\frac{1}{f}=\frac{1}{20}+\frac{1}{-40}=\frac{2-1}{40}=\frac{1}{40}$ ⇒ f = +40 cm; P = +2.5 D

(b) Image with u = −60 cm: $\frac{1}{v}=\frac{1}{40}+\frac{1}{-60}=\frac{3-2}{120}=\frac{1}{120}$ ⇒ v = +120 cm

(c) Nature: Real (v is positive for lens with Cartesian convention ⇒ image is on opposite side of light propagation — real).

(a) Critical angle at glass-water interface: $\sin C = \frac{n_2}{n_1} = \frac{1.33}{1.5} = 0.887$ ⇒ C ≈ 62.5°

(b) Angle of refraction (i = 30°): $n_1\sin i = n_2\sin r$ ⇒ $1.5\times 0.5 = 1.33\times\sin r$ ⇒ $\sin r = 0.564$ ⇒ r ≈ 34.4°

(c) If i > 62.5°, Total Internal Reflection (TIR) occurs — all light reflects back into glass, none passes into water.

(a) Speed in glass: $v_g = \frac{c}{n_g} = \frac{3\times10^8}{1.5} = 2\times10^8$ m/s

(b) Speed in water: $v_w = \frac{c}{n_w} = \frac{3\times10^8}{1.33} = 2.26\times10^8$ m/s

(c) n (glass w.r.t. water): $n_{gw} = \frac{n_g}{n_w} = \frac{1.5}{1.33} = 1.128$

(a) Corrective lens: Object at ∞, image at −1.5 m (far point). $f = -1.5$ m ⇒ P = $-\frac{1}{1.5} = -0.67$ D (concave lens)

Lens formula: u = −25 cm (object at 25 cm), v = −60 cm (image at near point)

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}=\frac{1}{-60}-\frac{1}{-25}=\frac{-5+12}{300}=\frac{7}{300}$... actually: $\frac{1}{f}=\frac{1}{-60}+\frac{1}{25}$ — wait, using: $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ with v=−60, u=−25: $\frac{1}{f}=\frac{1}{-60}-\frac{1}{-25}=\frac{-5+12}{300}=\frac{7}{300}$ ⇒ f ≈ +42.9 cm

Correct approach: $P=\frac{1}{f(m)}=\frac{1}{0.429}\approx +2.33$ D (convex lens)

Verification: $\frac{1}{v}=\frac{1}{42.9}+\frac{1}{-25}=\frac{1}{42.9}-\frac{1}{25}=0.0233-0.04=-0.0167$ ⇒ v = −60 cm ✓

(a) Object at 2f (u = −24 cm): $\frac{1}{v}=\frac{1}{12}+\frac{1}{-24}=\frac{2-1}{24}=\frac{1}{24}$ ⇒ v = +24 cm; m = −1 (real, inverted, same size)

(b) Object at 1.5f (u = −18 cm): $\frac{1}{v}=\frac{1}{12}+\frac{1}{-18}=\frac{3-2}{36}=\frac{1}{36}$ ⇒ v = +36 cm; m = 36/(-18) = −2 (real, inverted, magnified ×2)

(a) Snell’s law: $\sin r = \frac{\sin 50°}{1.6}=\frac{0.766}{1.6}=0.479$ ⇒ r ≈ 28.6°

(b) Lateral shift: $d = \frac{t\sin(i-r)}{\cos r} = \frac{6\times\sin(21.4°)}{\cos(28.6°)} = \frac{6\times 0.365}{0.877} = \frac{2.19}{0.877} \approx \mathbf{2.50}$ cm

(c) Incident and emergent rays are parallel (same direction) but laterally displaced by 2.5 cm. The angle between them = 0°.

(a) Combination focal length: $\frac{1}{f}=\frac{1}{10}+\frac{1}{-30}=\frac{3-1}{30}=\frac{2}{30}$ ⇒ f = +15 cm

(b) Image distance: $\frac{1}{v}=\frac{1}{15}+\frac{1}{-15}=0$ ⇒ v = ∞

(c) Magnification: Image at infinity; object is at focal point of combination. m = ∞ (object exactly at F).

(b) Magnifying glass: Object at F (40 cm from concave lens): image at ∞ ⇒ m = D/f = 25/40 = 0.625 (less than 1, so this concave lens is NOT used as magnifier)

(c) For virtual image at 40 cm: v = −40 cm; $\frac{1}{u}=\frac{1}{v}-\frac{1}{f}=\frac{1}{-40}-\frac{1}{-40}=0$ ⇒ This gives u = ∞. For finite image: $\frac{1}{u}=\frac{1}{-40}-\frac{1}{-40}$... use f=−40, v=−40: $\frac{1}{u}=\frac{1}{-40}+\frac{1}{40}=0$ ⇒ u=∞. Object at infinity gives image at 40 cm.

(a) Image distance: $\frac{1}{v}=\frac{1}{5}+\frac{1}{-6}=\frac{6-5}{30}=\frac{1}{30}$ ⇒ v = +30 cm

(b) Magnification: $m=\frac{v}{u}=\frac{30}{-6}=-5$ (real, inverted, ×5 magnified)

(c) Projected image height: h′ = |m| × h = 5 × 2 = 10 cm

At minimum deviation: $n = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

$\sqrt{3} = \frac{\sin\left(\frac{60+\delta_m}{2}\right)}{\sin 30°} = \frac{\sin\left(30+\frac{\delta_m}{2}\right)}{0.5}$

$\sin\left(30+\frac{\delta_m}{2}\right) = 0.5\sqrt{3} = 0.866$ ⇒ $30+\frac{\delta_m}{2}=60°$ ⇒ $\delta_m = 60°$

Angle of incidence at min deviation: $i = \frac{A+\delta_m}{2} = \frac{60+60}{2} = \mathbf{60°}$

(i) Camera image height: u=−500cm, f=+5cm. $\frac{1}{v}=\frac{1}{5}+\frac{1}{-500}=\frac{100-1}{500}=\frac{99}{500}$ ⇒ v≈5.05cm. m=5.05/500=0.0101; h′=0.0101×200=2.02cm on sensor.

(ii) Eye parts: (a) Lens → Crystalline lens; (b) Iris/diaphragm → Iris (controls pupil size); (c) Retina → Retina (film/sensor); (d) Ciliary muscles → Ciliary muscles (change focal length).

(iii) Power of accommodation: The ability of the eye to change the focal length of the crystalline lens (by ciliary muscle action) to focus objects at different distances. Least distance of distinct vision (D) = 25 cm (normal eye near point); Far point = infinity (normal). Near point is closest clear vision point; far point is the farthest.

(i) Defects: Patient X: Hypermetropia (near point too far); Patient Y: Myopia (far point too close); Patient Z: Presbyopia (age-related loss of accommodation).

(ii) Patient X (near point 80cm, need 25cm): u=−25cm, v=−80cm (image at near point). $P=\frac{1}{v}-\frac{1}{u}=\frac{1}{-0.80}-\frac{1}{-0.25}=-1.25+4=+2.75$ D (convex)

(iii) Patient Y (far point 50cm): f=−50cm=−0.5m; P=−2 D (concave)

(iv) Presbyopia: Age-related hardening of the eye lens and weakening of ciliary muscles; cannot accommodate near or far. Corrected with bi-focal lenses (upper concave for far, lower convex for near).

(i) Critical angle ($C$):
$\sin C = \dfrac{n_2}{n_1} = \dfrac{1.52}{1.62} \approx 0.9383$
$C = \sin^{-1}(0.9383) \approx \mathbf{69.7^\circ}$ (or $\approx 69.6^\circ$).

(ii) Ray incident at $70^\circ$:
Since angle of incidence $i = 70^\circ > C \ (69.7^\circ)$, and the light travels from denser core ($n=1.62$) towards rarer cladding ($n=1.52$), yes, the ray will undergo Total Internal Reflection (TIR) completely into the core without escaping into the cladding.

(iii) Working Principle & Optical Fibre Diagram:
An optical fibre consists of an inner core of higher refractive index surrounded by an outer cladding of lower refractive index. Light entering the core at one end hits the core-cladding boundary at an angle greater than the critical angle ($i > C$). It undergoes successive total internal reflections along the entire length of the cable with virtually zero loss of energy, carrying optical signals at light speed.

(i) Twinkling of Stars: Stars are extremely distant point sources of light. As starlight passes through the continuously moving, turbulent layers of the atmosphere, its optical path and apparent brightness fluctuate randomly from moment to moment. This continuous fluctuation causes twinkling. Planets do not twinkle because they are much closer and act as extended collections of point sources; their brightness variations average out to zero.

(ii) Advance Sunrise and Delayed Sunset: When the Sun is slightly below the actual horizon, light from the Sun traveling through space enters progressively denser layers of the atmosphere and bends downwards towards the observer. Consequently, the Sun appears above the horizon about 2 minutes before actual sunrise and remains visible for about 2 minutes after actual sunset, lengthening the apparent day by ~4 minutes.

Case 1 — Object at Principal Focus ($u = -f = -15\text{ cm}$):
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \implies \dfrac{1}{v} = \dfrac{1}{-15} - \dfrac{1}{-15} = 0 \implies v = \mathbf{\infty}$
The reflected rays are parallel; the image is formed at infinity (real, inverted, highly enlarged).

Case 2 — Object moved 5 cm closer to mirror:
New object distance $u' = -(15 - 5) = \mathbf{-10\text{ cm}}$, $f = -15\text{ cm}$
$\dfrac{1}{v'} = \dfrac{1}{f} - \dfrac{1}{u'} = \dfrac{1}{-15} - \dfrac{1}{-10} = -\dfrac{1}{15} + \dfrac{1}{10} = \dfrac{-2 + 3}{30} = \dfrac{+1}{30}$
$v' = \mathbf{+30\text{ cm}}$
Magnification $m = -\dfrac{v'}{u'} = -\dfrac{+30}{-10} = \mathbf{+3}$
Nature: Formed 30 cm behind the mirror; virtual, erect, and magnified 3 times.

Given: Power of first lens $P_1 = +5\text{ D}$ (convex lens), Net combination power $P = +3\text{ D}$
Formula: $P = P_1 + P_2 \implies P_2 = P - P_1 = +3 - (+5) = \mathbf{-2\text{ D}}$
Focal length of second lens: $f_2 = \dfrac{1}{P_2} = \dfrac{1}{-2\text{ D}} = -0.5\text{ m} = \mathbf{-50\text{ cm}}$
Nature of second lens: Negative power and focal length indicate it is a diverging (concave) lens of focal length 50 cm.

(i) Maximum Magnification: Maximum magnification occurs when the virtual image is formed at the least distance of distinct vision ($v = -D = -25\text{ cm}$):
$m_{max} = 1 + \dfrac{D}{f} = 1 + \dfrac{25}{5} = 1 + 5 = \mathbf{6\times}$

(ii) Position of object for maximum magnification:
Using lens formula with $v = -25\text{ cm}$ and $f = +5\text{ cm}$:
$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \implies \dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f} = \dfrac{1}{-25} - \dfrac{1}{5} = \dfrac{-1 - 5}{25} = \dfrac{-6}{25}$
$u = -\dfrac{25}{6} \approx \mathbf{-4.17\text{ cm}}$
The object must be placed 4.17 cm in front of the lens (inside the focal point $F$).

(i) Dispersion of Light by Glass Prism: White light enters the prism, refracts and disperses at the first surface, reflects across the base, and refracts again upon emerging into air as a full spectrum (VIBGYOR).

(ii) Rainbow Formation by Raindrop: Sunlight enters a spherical suspended raindrop, undergoes refraction with dispersion at entry, internal reflection at the back curved wall, and refraction as it exits the droplet into the observer's eye.

In a primary rainbow:

• Red appears on the outer rim (top arc), emerging at an angle of approximately $42^\circ$ relative to the incoming sunlight ray.
• Violet appears on the inner rim (bottom arc), emerging at an angle of approximately $40^\circ$ relative to the incoming sunlight.

Physical Reason: Violet light has shorter wavelength, travels slower in water, and has higher refractive index than red light. Thus violet undergoes maximum total deviation inside each droplet. Droplets situated higher in the sky direct their $42^\circ$ red beams into the observer's eye, whereas droplets situated lower direct their $40^\circ$ violet beams into the eye. Hence, red appears at the top and violet at the bottom.

1. $\Delta ABO$ and $\Delta A'B'O$ are similar:
$\dfrac{A'B'}{AB} = \dfrac{OB'}{OB}$   — (1)

2. $\Delta MOF_2$ (where $MO = AB$ is perpendicular from lens top) and $\Delta A'B'F_2$ are similar:
$\dfrac{A'B'}{MO} = \dfrac{A'B'}{AB} = \dfrac{F_2B'}{OF_2} = \dfrac{OB' - OF_2}{OF_2}$   — (2)

3. Equating (1) and (2): $\dfrac{OB'}{OB} = \dfrac{OB' - OF_2}{OF_2}$
Using Cartesian sign convention: $OB = -u$, $OB' = +v$, $OF_2 = +f$
$\dfrac{+v}{-u} = \dfrac{v - f}{f} \implies vf = -u(v - f) \implies vf = -uv + uf$
Dividing both sides throughout by $uvf$:
$\dfrac{vf}{uvf} = -\dfrac{uv}{uvf} + \dfrac{uf}{uvf} \implies \dfrac{1}{u} = -\dfrac{1}{f} + \dfrac{1}{v}$
Rearranging gives the standard lens formula: $\boxed{\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}}$

Given: $h = 3\text{ cm}$, $u = -8\text{ cm}$, $f = +12\text{ cm}$
$\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{12} + \dfrac{1}{-8} = \dfrac{2 - 3}{24} = -\dfrac{1}{24} \implies v = \mathbf{-24\text{ cm}}$
Image distance: 24 cm in front of lens (same side as object).
Magnification: $m = \dfrac{v}{u} = \dfrac{-24}{-8} = \mathbf{+3}$
Image height: $h' = m \times h = (+3) \times 3\text{ cm} = \mathbf{+9\text{ cm}}$
Nature: Virtual, erect, and magnified ($\times 3$), formed 24 cm from the optical centre.

A bi-focal lens corrects presbyopia (often accompanied by myopia):

• Upper half (Concave lens, $f = -2\text{ m}$, $P = -0.5\text{ D}$): Used for distant vision (myopia correction), allowing the wearer to see remote objects clearly.
• Lower half (Convex lens, $f = +40\text{ cm} = +0.4\text{ m}$, $P = +2.5\text{ D}$): Used for near vision (reading and close tasks), compensating for the loss of accommodation power of the ciliary muscles.

(a) Critical Angle calculation:
$\sin C = \dfrac{1}{n} = \dfrac{1}{1.52} \approx 0.6579 \implies C = \sin^{-1}(0.6579) \approx \mathbf{41.1^\circ}$

(b) Deviation by $90^\circ$ and $180^\circ$:
Since the critical angle of crown glass ($41.1^\circ$) is less than $45^\circ$, any normal ray striking an interior face at $45^\circ$ undergoes Total Internal Reflection ($i = 45^\circ > C$).
Porro Prism (Periscope/Binoculars): Light strikes the hypotenuse at $45^\circ$, completely reflecting by $90^\circ$ at the first internal face and $90^\circ$ at the second, turning the ray through a total of $180^\circ$ without loss of brightness. Unlike silvered mirrors which absorb 5-10% of light, TIR provides 100% reflection efficiency.

(a) Rayleigh Scattering Law:
Lord Rayleigh showed that for particles much smaller than the wavelength of light ($\text{size} \ll \lambda$), the intensity of scattered light $I$ is inversely proportional to the fourth power of wavelength:
$I \propto \dfrac{1}{\lambda^4}$
Since blue light has approximately half the wavelength of red light ($\lambda_{blue} \approx 400\text{ nm}$, $\lambda_{red} \approx 700\text{ nm}$):
$\dfrac{I_{blue}}{I_{red}} \approx \left(\dfrac{700}{400}\right)^4 \approx (1.75)^4 \approx \mathbf{9.4\text{ to }10\text{ times}}$ more intensely scattered than red.

(b) Red Appearance of Sun at Dawn and Dusk:
At sunrise and sunset, sunlight traverses the thickest layer of the atmosphere. During this long optical path, shorter blue and violet wavelengths are almost completely scattered away in all directions away from the observer's line of sight. Only the least-scattered longer wavelengths (red, orange) penetrate through directly to reach the observer's eye, making the Sun appear reddish.

(c) Appearance of Sky to an Astronaut:
In outer space, there is no atmosphere and no air molecules or dust particles to scatter sunlight. With no scattered light entering the astronaut's eyes from empty space, the sky appears completely dark and black, and stars are visible clearly even during daytime.

(a) Sign Convention Rules (New Cartesian):
1. All distances are measured from the optical centre ($O$) of a lens or pole ($P$) of a mirror.
2. Distances measured in the direction of incident light (along positive x-axis) are taken as positive.
3. Distances measured opposite to the direction of incident light are taken as negative.
4. Heights measured perpendicularly above the principal axis are positive; downwards are negative.

(b) Lens Immersion in Different Liquids:
The focal length of a glass lens ($n_g = 1.5$) in a liquid of refractive index $n_l$ is given by the Lens Maker's Formula: $\dfrac{1}{f_{med}} = \left(\dfrac{n_g}{n_l} - 1\right)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$.

• In water ($n_l = 1.33 < n_g$): $\left(\dfrac{1.5}{1.33} - 1\right) \approx 0.128$, which is smaller than in air ($1.5 - 1 = 0.5$). Thus focal length increases by a factor of ~4 ($f_{water} \approx 4 f_{air}$). Converging nature is preserved.
• In liquid with $n_l = 1.5 = n_g$: $\dfrac{n_g}{n_l} - 1 = 0 \implies f = \infty$. The lens becomes optically invisible and behaves as a plain glass plate.
• In liquid with $n_l = 1.65 > n_g$ (e.g. carbon disulfide): $\left(\dfrac{1.5}{1.65} - 1\right) < 0$. The sign of focal length reverses! A convex lens behaves as a diverging (concave) lens.

When light passes through a rectangular glass slab with parallel faces, refraction at the first surface is exactly compensated by refraction at the second parallel surface: $\sin i_1 = n \sin r_1$ and $n \sin r_2 = \sin e$. Because alternate interior angles $r_1 = r_2$, the emergent angle $e = i$.
The emergent ray is parallel to the incident ray but shifted sideways by perpendicular distance $d = \dfrac{t \sin(i - r)}{\cos r}$.

A ray coming from the bottom of a pool of real depth $h = 2\text{ m}$ ($n_w = \frac{4}{3}$) bends away from the normal upon exiting into air.
$\text{Apparent depth } h' = \dfrac{\text{Real depth } h}{n} = \dfrac{2\text{ m}}{4/3} = \mathbf{1.5\text{ m}}$
The bottom appears raised by $\Delta h = h - h' = 2 - 1.5 = \mathbf{0.5\text{ m}}$ (or 50 cm).

The crystalline lens is flexible and held in place by ciliary muscles:

• Viewing distant objects ($u \to \infty$): Ciliary muscles relax → suspensory ligaments tighten → lens becomes thin → radius of curvature increases → focal length increases to its maximum (~2.5 cm, distance from lens to retina).
• Viewing nearby objects ($u = 25\text{ cm}$): Ciliary muscles contract → suspensory ligaments slacken → lens rounds up and becomes thick → curvature increases → focal length decreases, ensuring sharp focus on retina.

In a hypermetropic eye, the eyeball is too short or focal length of the eye lens is too long. Rays from the standard near point ($25\text{ cm}$) focus behind the retina. A convex corrective lens provides initial convergence so that incoming rays appear to originate from the patient's defective near point ($N'$), focusing accurately on the retina.

(a) Resistance: $R = \frac{\rho l}{A} = \frac{1.0\times10^{-6}\times 1.5}{1.5\times10^{-6}} = \mathbf{1\,\Omega}$

(b) Current at V=6V: $I = \frac{V}{R} = \frac{6}{1} = \mathbf{6\,A}$

(c) Power: $P = VI = 6\times 6 = \mathbf{36\,W}$ (or $P=I^2R=36\times1=36$ W)

(a) Equivalent resistance: $\frac{1}{R_{eq}}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=\frac{3+2+1}{12}=\frac{6}{12}=\frac{1}{2}$ ⇒ $R_{eq}=\mathbf{2\,\Omega}$

(b) Current through each: $I_1=12/4=\mathbf{3\,A}$; $I_2=12/6=\mathbf{2\,A}$; $I_3=12/12=\mathbf{1\,A}$

(c) Total current: $I=3+2+1=\mathbf{6\,A}$ (or $I=12/2=6$ A) ✓

(d) Power: $P_1=I_1^2R_1=9\times4=36$ W; $P_2=4\times6=24$ W; $P_3=1\times12=12$ W; $P_{total}=72$ W

(a) Parallel (6∥3): $R_p = \frac{6\times3}{6+3}=\frac{18}{9}=2\,\Omega$; $R_{eq}=4+2=\mathbf{6\,\Omega}$

(b) Total current: $I=12/6=\mathbf{2\,A}$

(c) Voltage across parallel (A to B): $V_{AB}=I\times R_p=2\times2=4$ V; $I_{6\Omega}=4/6=0.67$ A; $I_{3\Omega}=4/3=1.33$ A

(d) Voltage across 4Ω: $V=2\times4=\mathbf{8\,V}$

(e) Power in parallel combination: $P_p=V_{AB}\times I=4\times2=\mathbf{8\,W}$

$l'\!=4l$, $A'\!=A/4$. $R'\!=\rho\frac{4l}{A/4}=\rho\frac{16l}{A}=16R=16\times8=\mathbf{128\,\Omega}$

(b) Original: V=IR=3×8=24V. New: I′=24/128=0.1875A≈0.19A

(c) Power ratio: $\frac{P_{before}}{P_{after}}=\frac{I^2R}{I'^2R'}=\frac{9\times8}{0.1875^2\times128}=\frac{72}{4.5}=\mathbf{16}$

Series: $R_s=10\,\Omega$; $I_s=20/10=2\,A$; $P_{series}=I^2R_s=4\times10=40$ W

Parallel: $R_p=\frac{6\times4}{10}=2.4\,\Omega$; $I_p=20/2.4=8.33$ A; $P_{parallel}=20\times8.33=166.7$ W

(a) Series current = 2A; Parallel total = 8.33A

(b) Current through P in series = 2A; in parallel = 20/6 = 3.33A

(c) Power: series=40W, parallel=166.7W

(d) Q in parallel: $P_Q=V^2/R_Q=400/4=100$ W; Q in series: $P_Q=I^2\times4=16$ W. Q consumes more power in parallel.

(a) $R_p=\frac{4\times6}{10}=2.4\,\Omega$; $R_{eq}=10+2.4=\mathbf{12.4\,\Omega}$

(b) $I_{total}=40/12.4=\mathbf{3.23\,A}$

(c) $V_{parallel}=I\times R_p=3.23\times2.4=7.74$ V; $I_{heater}=7.74/4=\mathbf{1.94\,A}$; $I_{lamp}=7.74/6=\mathbf{1.29\,A}$

(d) Voltage across parallel: $\mathbf{7.74\,V}$

(a) $R=V^2/P=220^2/840=48400/840=\mathbf{57.6\,\Omega}$

(b) $I=P/V=840/220=\mathbf{3.82\,A}$

(c) Heat in 30 min: $H=Pt=840\times30\times60=\mathbf{1.512\times10^6\,J}$

(d) Monthly cost: Energy/day = 840W × 0.5h = 0.42 kWh; Monthly = 0.42×30=12.6 kWh; Cost=12.6×6=₹75.6

(a) $R=V^2/P=250^2/100=\mathbf{625\,\Omega}$

(b) $I=V/R=200/625=\mathbf{0.32\,A}$

(c) Actual power: $P_{actual}=I^2R=0.32^2\times625=0.1024\times625=\mathbf{64\,W}$

(d) Ratio: $P_{actual}/P_{rated}=64/100=\mathbf{0.64}$ (only 64% of rated power)

(a) Parallel (220V each): Each gets full 220V ⇒ each consumes 60W

(b) Series: $R_{total}=3\times806.7=2420\,\Omega$; $I=220/2420=0.091\,A$; Power each = $I^2R=0.091^2\times806.7=\mathbf{6.67\,W}$

(c) Bulbs glow much brighter in parallel (60W each vs 6.67W each)

(d) Ratio: $P_{parallel}/P_{series}=3\times60/(3\times6.67)=180/20=\mathbf{9}$

(a) $I=E/(R+r)=12/12=\mathbf{1\,A}$

(b) Terminal voltage: $V_T=E-Ir=12-1\times2=\mathbf{10\,V}$

(c) Power to external: $P_R=I^2R=1\times10=\mathbf{10\,W}$

(d) Power wasted in r: $P_r=I^2r=1\times2=\mathbf{2\,W}$; Efficiency=$\frac{P_R}{P_{total}}=\frac{10}{12}=\mathbf{83.3\%}$

Refrigerator: 0.2×24=4.8 kWh; 2 Fans: 2×0.075×12=1.8; 3 LEDs: 3×0.01×8=0.24; TV: 0.12×5=0.6; Iron: 1×1=1; Geyser: 2×0.5=1

(a) Total/day: 4.8+1.8+0.24+0.6+1+1=9.44 kWh

(b) Monthly bill: 9.44×30×7=₹1982.40

(c) Annual savings (no geyser): Geyser/day=1 kWh; Annual=365 kWh; Savings=365×7=₹2555/year

(a) At 200°C: $R=10(1+0.0004\times200)=10(1.08)=\mathbf{10.8\,\Omega}$

(b) R=12Ω: $12=10(1+0.0004t)$ ⇒ $1.2=1+0.0004t$ ⇒ $t=0.2/0.0004=\mathbf{500°C}$

(c) 0 to 500°C: $R_{500}=10(1+0.0004\times500)=10\times1.2=12\,\Omega$; $\%$ increase $=\frac{12-10}{10}\times100=\mathbf{20\%}$

(a) R₂∥R₃: $R_{23}=\frac{6\times3}{9}=2\,\Omega$; $R_{eq}=2+2+4=\mathbf{8\,\Omega}$

(b) $I=20/8=\mathbf{2.5\,A}$

(c) V across parallel: $V_{23}=2.5\times2=5$ V

(d) $I_{R2}=5/6=0.833$ A; $I_{R3}=5/3=1.667$ A

(e) Power in R₄: $P_4=I^2R_4=2.5^2\times4=6.25\times4=\mathbf{25\,W}$

(a) Total current: $I=\frac{100+40+1200}{220}=\frac{1340}{220}=\mathbf{6.09\,A}$

(b) 6.09A > 5A ⇒ MCB trips. The hair dryer (1200W) is the main culprit (draws 5.45A alone).

(c) Solution: Switch off the hair dryer before using fan & bulb together. Or request a higher-rated MCB (10A) or use dryer on a separate circuit.

(d) Monthly bill: Energy/day=(0.1×8)+(0.04×5)+(1.2×0.25)=0.8+0.2+0.3=1.3 kWh; Monthly=1.3×30×8=₹312

(a) Resistances: $R_{60}=220^2/60=806.7\,\Omega$; $R_{100}=220^2/100=484\,\Omega$

(b) Series current: $I=440/(806.7+484)=440/1290.7=\mathbf{0.341\,A}$

(c) Power: $P_{60}=0.341^2\times806.7=\mathbf{93.8\,W}$; $P_{100}=0.341^2\times484=\mathbf{56.3\,W}$

(d) The 60W bulb is more likely to fuse because it has higher resistance and hence consumes MORE power (93.8W > 56.3W) than its rating (60W).

(a) R: Incandescent: $R=220^2/60=806.7\,\Omega$; LED: $R=220^2/9=5377.8\,\Omega$

(b) 8 incandescent bulbs, 6h/day: Energy = 8×0.06×6 = 2.88 kWh/day

(c) 8 LEDs, 6h/day: Energy = 8×0.009×6 = 0.432 kWh/day

(d) Annual savings: Saved/day = 2.88−0.432 = 2.448 kWh; Annual = 2.448×365 = 893.5 kWh; Money saved = 893.5×5 = ₹4467.6/year

(i) $I=V/R=220/44=\mathbf{5\,A}$

(ii) $H=I^2Rt=25\times44\times600=\mathbf{660,000\,J}=660\,kJ$

(iii) $H=mc\Delta T$ ⇒ $660000=0.5\times c\times(100-30)=0.5\times c\times70=35c$ ⇒ $c=660000/35=\mathbf{18857\,J/kg°C}$

⚠ This is higher than 4200 J/kg°C, meaning some heat is lost. If all 660kJ went to water: $c=660000/(0.5\times70)=18857$ J/kg°C. Real water c=4200, so efficiency=4200/18857≈22% — indicating 78% heat lost to environment.

(i) Electrical input: $P_{in}=VI=220\times10=\mathbf{2200\,W}$

(ii) Mechanical output: $P_{mech}=0.80\times2200=\mathbf{1760\,W}$

(iii) Heat/hour: $P_{heat}=0.20\times2200=440$ W; $H=440\times3600=\mathbf{1.584\times10^6\,J}$

(iv) Cost for 8h: Energy = 2.2×8 = 17.6 kWh; Cost = 17.6×6 = ₹105.6

(i) Total power: 10×75 + 40×18 + 5×300 + 750 = 750+720+1500+750 = 3720 W = 3.72 kW

(ii) Total current at 230V: $I=3720/230=\mathbf{16.17\,A}$

(iii) Annual energy: 3.72×8×200 = 5952 kWh/year

(iv) Panels needed: 3720/250 = 14.88 ⇒ 15 panels (round up)

(i) Total current: $I=(200+150+1500+75+75)/220=2000/220=9.09\,A$ < 15A ⇒ MCB does NOT trip.

(ii) (a) Overloading: When total current exceeds safe limit due to too many appliances; wires heat up ⇒ fire risk.
(b) Short circuit: Live and neutral wires accidentally touch (zero resistance path) ⇒ huge current ⇒ fire.
(c) Earth wire: Connected to metal body of appliances. If live wire touches body, current flows safely to earth instead of through user ⇒ prevents shock.

(iii) Monthly bill (8h/day, 30 days): Energy = 2.0×8×30=480 kWh; Cost=480×7=₹3360

(i) B∥C: $R_{BC}=\frac{3\times6}{9}=2\,\Omega$; $R_{eq}=A+R_{BC}+D=2+2+1=5\,\Omega$ (external); total=5+1=6Ω

(ii) $I=12/6=\mathbf{2\,A}$

(iii) V across B (=V across C): $V_{BC}=I\times R_{BC}=2\times2=4\,V$; $I_B=4/3=\mathbf{1.33\,A}$

(iv) Power in C: $P_C=V_{BC}^2/R_C=16/6=\mathbf{2.67\,W}$

(v) Terminal voltage: $V_T=E-Ir=12-2\times1=\mathbf{10\,V}$

$R_{eq}=V/I=6/6=1\,\Omega$

$\frac{1}{R_Z}=\frac{1}{R_{eq}}-\frac{1}{R_X}-\frac{1}{R_Y}=1-\frac{1}{2}-\frac{1}{3}=\frac{6-3-2}{6}=\frac{1}{6}$ ⇒ $R_Z=6\,\Omega$

Power in Z: $P_Z=V^2/R_Z=36/6=\mathbf{6\,W}$

$A=\pi r^2=3.14\times(3\times10^{-4})^2=3.14\times9\times10^{-8}=2.83\times10^{-7}\,m^2$

$R=\rho l/A$ ⇒ $l=\frac{RA}{\rho}=\frac{30\times2.83\times10^{-7}}{1.2\times10^{-6}}=\frac{8.49\times10^{-6}}{1.2\times10^{-6}}=\mathbf{7.07\,m}$

Why nichrome in heaters: High resistivity ⇒ high heat per unit length; high melting point; does not oxidise. NOT used in transmission: very high resistivity would waste enormous energy as heat over long distances.

Case A (Series): EMF=6V, r=1.5Ω; $I_A=\frac{6}{3+1.5}=\frac{6}{4.5}=\mathbf{1.33\,A}$; $V_T=6-1.33\times1.5=\mathbf{4\,V}$; $P_R=1.33^2\times3=\mathbf{5.33\,W}$

Case B (Parallel): EMF=2V, r=0.5/3=0.167Ω; $I_B=\frac{2}{3+0.167}=\frac{2}{3.167}=\mathbf{0.632\,A}$; $V_T=2-0.632\times0.167=\mathbf{1.89\,V}$; $P_R=0.632^2\times3=\mathbf{1.20\,W}$

Each battery supplies I+/3 = 0.632/3 = 0.21A in parallel; In series each supplies 1.33A. Parallel preferred for high-current, low-resistance loads to avoid large voltage drop across internal resistance.

1. Independent Operation: Each appliance has its own dedicated switch. Turning one appliance on or off does not affect the operation of others. In a series circuit, if one appliance is turned off or breaks down, the whole circuit opens and every appliance stops working.

2. Constant Full Rated Voltage (220 V): In parallel, each appliance receives the full mains supply potential difference of 220 V, operating at its designed power rating. In series, voltage divides across appliances according to their resistances, causing bulbs to glow very dimly.

3. Low Total Resistance & High Current Capacity: Connecting in parallel reduces the overall effective resistance of the circuit ($1/R_p = \sum 1/R_i$), allowing the circuit to draw adequate current from the supply to power heavy-load appliances (like heaters and air conditioners).

Given: Supply voltage $V = 220\text{ V}$, Fuse resistance $R_{fuse} = 0.1\ \Omega$, Short-circuit wire resistance $R_{wire} = 0.1\ \Omega$
Total resistance during short circuit: $R_{total} = R_{fuse} + R_{wire} = 0.1 + 0.1 = \mathbf{0.2\ \Omega}$
Instantaneous fault current: $I_{sc} = \dfrac{V}{R_{total}} = \dfrac{220\text{ V}}{0.2\ \Omega} = \mathbf{1100\text{ A}}$
This enormous surge of current heats the fuse wire rapidly ($H = I^2 R t$), causing it to melt within milliseconds and safely interrupt the circuit before fire breaks out.

(i) Total energy stored in battery pack:
$E = V \times Q = 48\text{ V} \times 100\text{ Ah} = 4800\text{ Wh} = \mathbf{4.8\text{ kWh}}$ (or $4.8 \times 3.6 \times 10^6\text{ J} = 1.728 \times 10^7\text{ J}$).

(ii) Driving range (time) on full charge:
$t = \dfrac{\text{Capacity}}{\text{Current drawn}} = \dfrac{100\text{ Ah}}{20\text{ A}} = \mathbf{5\text{ hours}}$ of continuous driving.

(iii) Specifications of each cell (16 identical cells in series):
In series connection, voltages add up, but charge capacity remains identical:
$\text{EMF of each cell} = \dfrac{48\text{ V}}{16} = \mathbf{3.0\text{ V}}$
$\text{Capacity of each cell} = \mathbf{100\text{ Ah}}$ (same through series branch).

(iv) Charging energy and cost:
Charging power: $P_{charge} = V_{in} \times I_{in} = 50\text{ V} \times 15\text{ A} = 750\text{ W} = 0.75\text{ kW}$
Charging energy for 4 hours: $E_{charge} = 0.75\text{ kW} \times 4\text{ h} = \mathbf{3.0\text{ kWh}}$
Cost of one full charge at ₹8 per kWh: $\text{Cost} = 3.0 \times 8 = \mathbf{\text{\rupee}24}$.

(v) Heat generated in motor winding per hour:
Current $I = 20\text{ A}$, Winding resistance $R = 0.5\ \Omega$, Time $t = 1\text{ hour} = 3600\text{ s}$
$H = I^2 R t = (20)^2 \times 0.5 \times 3600 = 400 \times 0.5 \times 3600 = \mathbf{720,000\text{ J}} = \mathbf{720\text{ kJ}}$ per hour.