#glass slab

Refraction at Spherical Surfaces and Lens Maker's Formula

A glass slab (\( n = 1.6 \)) of thickness \( 9.6 \, \text{cm} \) is placed over a mark. What is the apparent shift?

**Refraction at spherical surface** formula n₁/u + n₂/v = (n₂-n₁)/R governs single surface, extension to two surfaces yields lens maker. Double convex with equal |R| has f = R/[2(n-1)], for R=12 cm, n=1.5, f=12 cm, illustrating dependence on curvature and index. Shift = t ( 1 - (1/n) ) . t = 9.6 cm , n = 1.6 . Shift = 9.6 ( 1 - (1/1.6) ) = 9.6 ( 1 - 0.625 ) = 9.6 × 0.375 = 3.6 cm . Substituting values gives 3.6 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f

Ref: NCERT > Physics Book > Ray Optics > Refraction at Spherical Surfaces and Lens Maker's Formula

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Refraction Through Prism and Minimum Deviation

A glass slab (\( n = 1.6 \)) of thickness \( 8 \, \text{cm} \) is placed over a point. What is the apparent shift?

**Prism minimum deviation** δ_m satisfies n = sin[(A+δ_m)/2]/sin(A/2), A prism angle (degrees), n refractive index, δ_m minimum deviation. For A=60°, n=1.45, sin[(60+δ_m)/2]=1.45×sin30°=0.725, (60+δ_m)/2=46.5°, 60+δ_m=93°, δ_m=33°, illustrating n increase raises δ_m. Shift = t ( 1 - (1/n) ) . t = 8 cm , n = 1.6 . Shift = 8 ( 1 - (1/1.6) ) = 8 ( 1 - 0.625 ) = 8 × 0.375 = 3 cm . Substituting values gives 3 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.

Ref: NCERT > Physics Book > Ray Optics > Refraction Through Prism and Minimum Deviation

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Reflection by Spherical Mirrors and Mirror Formula

A glass slab (\( n = 1.5 \)) of thickness \( 15 \, \text{cm} \) is placed over a pin. By how much does the pin appear ra

**Concave mirror image formation** depends on object position: beyond C real inverted diminished between F and C, at C real inverted same size at C, between C and F real inverted magnified beyond C, at F image at infinity, within F virtual erect magnified behind mirror. Shift = t ( 1 - (1/n) ) . t = 15 cm , n = 1.5 . Shift = 15 ( 1 - (1/1.5) ) = 15 ( 1 - (2/3) ) = 15 × (1/3) = 5 cm . Substituting values gives 5 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v

Ref: NCERT > Physics Book > Ray Optics > Reflection by Spherical Mirrors and Mirror Formula

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Reflection by Spherical Mirrors and Mirror Formula

A glass slab (\( n = 1.5 \)) of thickness \( 7.5 \, \text{cm} \) is placed over a point. What is the apparent shift?

**Spherical mirror reflection** follows law θ_i = θ_r, focal length f = R/2, concave R negative, convex positive. Object beyond center C (u > 2f) forms image between F and C diminished, at C same size, between C and F magnified, beyond F forms at infinity, illustrating mirror equation. Shift = t ( 1 - (1/n) ) . t = 7.5 cm , n = 1.5 . Shift = 7.5 ( 1 - (1/1.5) ) = 7.5 ( 1 - (2/3) ) = 7.5 × (1/3) = 2.5 cm . Substituting values gives 2.5 cm, which matches expected image position and magnification from mirror/lens formula

Ref: NCERT > Physics Book > Ray Optics > Reflection by Spherical Mirrors and Mirror Formula

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Refraction at Plane Surfaces and Snell's Law

A glass slab (\( n = 1.52 \)) of thickness \( 7.6 \, \text{cm} \) is placed over a dot. What is the apparent shift?

**Snell's law** n₁ sinθ₁ = n₂ sinθ₂ describes refraction at plane interface, n refractive index, θ angle with normal. When light goes from denser n=1.52 glass to rarer air n=1, sinθ₂ = (n₁/n₂) sinθ₁ > sinθ₁, bending away from normal, enabling total internal reflection beyond critical angle. Shift = t ( 1 - (1/n) ) . t = 7.6 cm , n = 1.52 . Shift = 7.6 ( 1 - (1/1.52) ) = 7.6 ( 1 - 0.658 ) = 7.6 × 0.342 ≈ 2.6 cm . Substituting values gives 2.6 cm, which matches expected image position and magnification from mirror/lens formula 1/f =

Ref: NCERT > Physics Book > Ray Optics > Refraction at Plane Surfaces and Snell's Law

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Reflection by Spherical Mirrors and Mirror Formula

A glass slab (\( n = 1.5 \)) of thickness \( 9 \, \text{cm} \) is placed over a mark. What is the apparent shift?

**Mirror formula** 1/f = 1/v + 1/u governs spherical mirrors, f = R/2, R radius of curvature (m), u object distance (m), v image distance (m), sign convention: distances in front of mirror negative for real is convention but magnitude used, magnification m = -v/u, concave forms real inverted when object beyond F, virtual erect within F. Shift = t ( 1 - (1/n) ) . t = 9 cm , n = 1.5 . Shift = 9 ( 1 - (1/1.5) ) = 9 ( 1 - (2/3) ) = 9 × (1/3) = 3 cm . Substituting values gives 3 cm, which matches

Ref: NCERT > Physics Book > Ray Optics > Reflection by Spherical Mirrors and Mirror Formula

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Refraction Through Prism and Minimum Deviation

A glass slab (\( n = 1.62 \)) of thickness \( 8.1 \, \text{cm} \) is placed over a dot. What is the apparent shift?

**Refraction through prism** deviation δ = i+e-A, minimum when i=e, symmetrical path, r₁=r₂=A/2, n = sin[(A+δ_m)/2]/sin(A/2). In water n_rel = n_prism/n_water =1.5/1.33=1.128, so δ_m reduces because relative index lower, new δ_m from formula with n_rel. Shift = t ( 1 - (1/n) ) . t = 8.1 cm , n = 1.62 . Shift = 8.1 ( 1 - (1/1.62) ) = 8.1 ( 1 - 0.617 ) = 8.1 × 0.383 ≈ 3.1 cm . Substituting values gives 3.1 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.

Ref: NCERT > Physics Book > Ray Optics > Refraction Through Prism and Minimum Deviation

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Reflection by Spherical Mirrors and Mirror Formula

A glass slab (\( n = 1.5 \)) of thickness \( 12 \, \text{cm} \) is placed over a mark. What is the apparent shift?

**Concave mirror image formation** depends on object position: beyond C real inverted diminished between F and C, at C real inverted same size at C, between C and F real inverted magnified beyond C, at F image at infinity, within F virtual erect magnified behind mirror. Shift = t ( 1 - (1/n) ) . t = 12 cm , n = 1.5 . Shift = 12 ( 1 - (1/1.5) ) = 12 ( 1 - (2/3) ) = 12 × (1/3) = 4 cm . Substituting values gives 4 cm, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v

Ref: NCERT > Physics Book > Ray Optics > Reflection by Spherical Mirrors and Mirror Formula

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Reflection by Spherical Mirrors and Mirror Formula

A glass slab (\( n = 1.5 \)) of thickness \( 6 \, \text{cm} \) is placed over a point. What is the apparent shift?

**Spherical mirror reflection** follows law θ_i = θ_r, focal length f = R/2, concave R negative, convex positive. Object beyond center C (u > 2f) forms image between F and C diminished, at C same size, between C and F magnified, beyond F forms at infinity, illustrating mirror equation. Shift = t ( 1 - (1/n) ) . t = 6 cm , n = 1.5 . Shift = 6 ( 1 - (1/1.5) ) = 6 ( 1 - (2/3) ) = 6 × (1/3) = 2 cm . Substituting values gives 2 cm, which matches expected image position and magnification from mirror/lens formula

Ref: NCERT > Physics Book > Ray Optics > Reflection by Spherical Mirrors and Mirror Formula

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Elastic Properties and Applications

A glass slab of volume 0.02m3 is subjected to a hydraulic pressure of 5×106N/m2. If the bulk modulus of glass is 3.7×101

Bulk modulus: B = −pΔVV. Rearrange: ΔVV = −pB = −5×1063.7×1010≈−1.35×10−4. Magnitude: 1.35×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.35×10−4. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Elastic Properties and Applications

A glass slab of volume 0.03m3 is subjected to a hydraulic pressure of 5×106N/m2. If the bulk modulus of glass is 3.7×101

Bulk modulus: B = −pΔVV. Rearrange: ΔVV = −pB = −5×1063.7×1010≈−1.35×10−4. Change in volume: ΔV = ΔVV×V = −1.35×10−4×0.03≈−4.05×10−6m3. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 4.05×10−6m3. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Elastic Properties and Applications

A glass slab of volume 0.04m3 is subjected to a hydraulic pressure of 7×106N/m2. If the bulk modulus of glass is 3.7×101

Bulk modulus: B = −pΔVV. Rearrange: ΔVV = −pB = −7×1063.7×1010≈−1.89×10−4. Magnitude: 1.89×10−4. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1.89×10−4. This satisfies dimensional consistency and physical conditions given, so option A is scientifically correct.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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