Title 09. RAY OPTICS AND OPTICAL INSTRUMENTS(H).pdf ✅

NEET REVISION 09. RAY OPTICS AND OPTICAL INSTRUMENTS(H) NEET REVISION Date: March 18, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on Object should appear to be at distance from mir‐ ror i.e., the apparent distance of the object w.r.to the mirror 2. () : Explana on According to displacement method . 3. () : Explana on The height of the image of the tower is given by 4. () : Explana on For a plane mirror the rela on between object and image veloci es w.r.to mirror is Consider that the mirror is at the origin and object is present on (-)ve -axis Given that 5. () : Explana on According to lens maker formula, Also, From eqn. (2) and (3), 6. () : Explana on By using lens makers formula. 7. () : Explana on When an object is placed infront of such a lens,the rays first of all refracted from the convex surface, then reflect from the polished plane surface and again refracts from convex surface. If and be the focal lengths of lens (convex surface) and mirror (plane polished surface) respec vely, then effec ve focal length is given by As, or R = R ⇒ μ ( ) + h = R d 1 ⇒ d = ( ) R−h μ O = √I1I2 O = √4 × 16 = 8 cm m = = ⇒ h2 = = cm f0 u0 h2 h1 f0h1 u0 14 3 VIm = −VOm x →V O = −6 ^i, Vm = 0 ⇒ → V1 = − →V O = +6 ^i V→ IO = V→ I − V→O = 6 ^i − (−6 ^i) = 12 m/s^i = ( ) ( − ) (1) 1 f ng−nm nm 1 R1 1 R2 Here, ang = = 1.5 ng na P = = 5D 1 f ∴ 5 = (1.5 − 1) ( − ) (2) 1 R1 1 R2 f = −100 cm = −1 m, nl =? ∴ = ( − 1) ( − ) (3) 1 −1 1.5 nl 1 R1 1 R2 −5 = ⇒ nl = 0.5 ( −1) 1.5 nl 5 3 R1 = 15 cm, R2 = −30 cm, f = 20 cm = (μ − 1) ( − ) 1 f 1 R1 1 R2 ⇒ = (μ − 1) ( + ) 1 20 1 15 1 30 = (μ − 1) ( ) ⇒ = (μ − 1) × ; 1 20 2+1 30 1 20 3 30 ⇒ μ = 1.5 fl fm F = + + 1 F 1 fl 1 fm 1 fl = + = (∵ fm = = ∞) 1 F 2 fl 1 fm 2 fl R 2 = (μ − 1) ( ) 1 fl 1 R ∴ = or F = 1 F 2(μ−1) R R 2(μ−1) R = 2F = R (μ−1)
NEET REVISION 8. () : Explana on Let me in object clock be me in image clock be then, 9. () : Explana on For no devia on, . 10. () : Explana on The given prism is a right angled prism and angle . Since, the ray is suffering total inter‐ nal reflec on and the cri cal angle is the angle of in‐ cidence in the denser medium for which the angle of redfrac on in the rarer medium is , and 11. () : Explana on As, is at a height above the mirror, image of will be at a depth below the mirror. If is depth of liquid in the tank, apparent depth of P, Apparent depth of image of is Apparent distance between and its image 12. () : Explana on When white light passes through a lens, violet (V)light undergoes more refrac on than red light (R). This discrepancy occurs because the wavelength of violet light is shorter than that of red light, resul ng in the focal length for red light being greater than that for violet light. So correct op on is (1). 13. () : Explana on So, 14. () : Explana on So, total of image is 3. 15. () : Explana on Since, grazes along surface angle cri cal angle Now, using snell's law on 1st interface, 16. () : Explana on The apparent depth will be sum of depth due water and oil 17. () : Explana on T1 & T2 T1 + T2 = 12 : 00 : 00 4 : 25 : 37 + T2 = 12 : 00 : 00 T2 = 07 : 34 : 23. A1 (μ1 − 1) = A2 (μ2 − 1) ⇒ A2 = (4 ∘ ) ( ) = 3 1 ∘ .54 − 1 1.72 − 1 ∠LMN = 45 ∘ 90 ∘ i. e. C = 45 ∘ μ = = = √2. 1 sin C 1 sin 45 ∘ P h P h d x1 = d−h μ P x2 = d+h μ ∴ P = x2 − x1 = − = d+h μ d−h μ 2h μ = (μ − 1) ( − ) 1 f 1 R1 1 R2 μR < μV Forobjective For eyepiece = = = − 1 f0 1 v0 1 u0 1 fe 1 ve 1 ue ⇒ = − ⇒ = − 1 1.5 1 v0 1 −2 1 6.25 1 −25 1 ve ⇒ = − ⇒ = − 1 v0 1 1.5 1 2 1 ue 1 −25 1 6.25 ⇒ v0 = 6 cm ⇒ ue = −5 cm L = |v0| + |ue| = (6 + 5)cm = 11 cm BC ⇒ (90 − θ1) = C ∴ sin C = 1 μ sin(90 − θ1) = ⇒ cos θ1 = √3 2 √3 2 ⇒ θ1 = 30 ∘ n1 sin θ = n2 sin θ1 ⇒ (1)sin θ = sin 30 ∘ = ⋅ 2 √3 2 √3 1 2 ⇒ sin θ = 1 √3 ⇒ θ = sin −1 ( ) 1 √3 = + = [ + ] = t1 μ1 t2 μ2 t 2 1 μ1 1 μ2 t(μ1 + μ2) 2μ1μ2 = ( − 1) [ − ] 1 f μ2 μ1 1 R1 1 R2 ⇒ = ( − 1) [ ] 1 1.2 9 8 2 R ⇒ R = 0.3m
NEET REVISION 18. () : Explana on Apparent depth real depth . From one side From opposite side Divide (1) by (2), we get 19. () : Explana on The len's maker formula is given as Apply lens maker formula in air Apply lens maker formula in liquid Also Dividing (1) by (2), we get. 20. () : Explana on When the final image is at the least distance of dis‐ nct vision, then When the final image is at infinity, then 21. () : Explana on The effec ve focal length is focal length of the mirror focal length of lens The lens will be like a concave mirror of focal length , i.e., nature will be converging. 22. () : Explana on For For So horizontal distance between image of and is 23. () : Explana on Given: For objec ve, For normal adjustment, the image formed by objec‐ ve will be at the focus of the eyepiece. 24. () : Explana on Consider the velocity component along normal direc‐ on =( )/μ x/μ = 12 cm (1) (24 − x)/μ = 4 cm (2) = ⇒ = 3 ⇒ x = 18 x/μ (24 − x)/μ 12 4 x 24 − x μ = = 18 12 3 2 = (mμg − 1) ( − ) 1 fa 1 R1 1 R2 3 = (1.25 − 1) ( − ) (1) 1 R1 1 R2 −2 = ( − 1) ( − ) (2) 1.25 μ 1 R1 1 R2 = ⇒ μ = 1.5 −3 2 0.25μ 1.25 − μ m = − (1 + ) f0 fe fe D = (1 + ) 200 5 5 25 = = −48 200 × 6 5 × 5 m = − = = −40 f0 fe 200 5 = − 1 F 1 fm 2 fl fm− fl− fm = R/2 = −20/2 = −10 cm 1/fl = ( − 1) [(1/R1) − (1/R2)] μ2 μ1 1/fl = [(2/3) − 1] [(1/20) − (1/ − 20] 1/fl = (−1/3)(1/10) = −1/30 1/F = (1/fm) − (2/fl) 1/F = (−1/10) − 2(−1/30) F = −30 cm 30 cm P : + = ⇒ + = ⇒ vP = − m 1 v 1 u 1 f 1 vP 1 −3 1 −1 3 2 Q : + = ⇒ vQ = m 1 vQ 1 −5 1 −1 −5 4 P Q 0.25 m fo = 50 m, fe = 10 m − = 1 vo 1 u0 1 fo ∴ − = 1 v0 1 −250 1 50 ∴ = − = = 1 v0 1 50 1 250 5 − 1 250 1 62.5 v0 = 62.5 m ∴ L = vo + fe = 72.5 m → v o = 4 ^j → v m = 0 → v om = 4 ^j ⇒ → v Im = −4 ^j ⇒ → v I = −4 ^j → v Io = → v I − → v o = −8 ^j
NEET REVISION 25. () : Explana on and For distant object, 26. () : Explana on 27. () : Explana on 28. () : Explana on Nega ve sign is for real, inverted image. Area enclosed by image 29. () : Explana on Using We get . 30. () : Explana on Apparent depth . 31. () : Explana on Distance of image from the plane surface is: For the curved side The minus sign means the image is virtual 32. () : Explana on The minimum length of the mirror required for the purpose is half the height of the person 33. () : Explana on As same lens, constant (say) Required It is an "increase" as it is posi ve. Correct op on is (2). 34. () : Explana on We have, Let and be the magnifica on when object is at and respec vely. Then, and As f0 = 140 cm fe = 5 cm m = = = 28 f0 fe 140 5 = − aμw f (aμg−1) R1 (aμg−aμw) R2 = − ( −1) 3 2 20 ( − ) 3 2 4 3 −20 = + = 1 40 1 120 1 30 ∴ f = × 30 = 40 cm 4 3 sin c = ⇒ = sin 45 ∘ = n2 n1 n2 n1 1 √2 ⇒ = √2 n1 n2 = − 1 v 1 f 1 u = − 1 −18 1 −90 = = − 1−5 90 2 45 ∴ v = − cm 45 2 m = = − h2 h1 v u ∴ = − = − h2 4.8 −45 2 −90 1 4 ∴ h2 = −1.2 cm ∴ = (1.2) 2 = 1.44 cm2 sin iC = = 1 μ r √r 2+h2 h = 12 cm, μ = 4/3 cm 36 √7 = + tb μb tw μw = + = (4 + 3)cm = 7 cm 6 cm 1.5 4 cm 4/3 x1 = = 2.5 cm 4 1.6 (dapp = ) dactual μ − = n2 v n1 u n2−n1 R n1 = 1.6, n2 = 1 R = 8cm + = 1.6 4 1 x2 1−1.6 8 ∴ x2 ≈ −3.0 cm ∴ I1I2 = (8 − 2.5 − 3.0) cm = 2.5 cm ,i. e. 0.8 m. [ − ] = 1 R1 1 R2 = K = [μa − 1] K 1 fa = [ − 1] K 1 fl μa μl ⇒ = = fl fa [1.5 − 1] [ − 1] 1.5 4/3 0.5 − 1 1.5 × 3 4 ⇒ = = 4 fl fa 0.5 × 4 0.5 ⇒ % = × 100 = [( ) − 1] × 100 fl − fa fa f1 f0 = (4 − 1) × 100 = 300% m = f u − f m m′ x y m = f −x − f m′ = f −y − f m′ = −m ⇒ = f −y − f −f −x − f ⇒ f = − ( ) x + y 2