Title 9.RAY OPTICS AND OPTICAL INSTRUMENTS - Explanations.pdf ✅

1 (a) I ∝ 1 r 2 ⇒ I2 I1 = r1 2 r2 2 = 602 1802 = 1 9 2 (a) Focal length in air is given by 1 faa = ( aμg − 1) ( 1 R1 − 1 R2 ) The focal length of lens immersed in water is given by 1 f1 = ( lng − 1) ( 1 R1 − 1 R2 ) When, R1, R2 are radii of curvatures of the two surfaces of lens and lng is refractive index of glass with respect to liquid. Also, lμg = ang a anl Given, a ang = 1.5, faa = 12 cm,a anl = 4 3 ∴ fl faa = (a ang − 1) ( lng − 1) f1 12 = (1.5 − 1) ( 1.5 4/3 − 1) = 0.5 × 4 0.5 ⇒ f1 = 4 × 12 = 48 cm 3 (a) Full use of resolving power means whole aperture of objective in use. And for relaxed vision fo fe = D d ⇒ 300 fe = 15 0.3 ⇒ fe = 6 cm 4 (b) m ∝ 1 fe 5 (c) 6 (a) When final image is formed at infinity, length of the tube = vo + fe ⇒ 15 = vo + 3 ⇒ vo = 12 cm For objective lens 1 fo = 1 vo − 1 uo ⇒ 1 (+2) = 1 (+12) − 1 u ⇒ uo = −2.4 cm 7 (a) For large objects, large image is formed on retina 8 (d) P = P1 + P2 = 100 f1 + 100 f2 , both f1 and f2 are in cm. = 100 50 − 100 40 =2-2.5=-0.5 D 9 (d) R.P. of microscope = 2μ sin θ λ 10 (a) Since light transmitting area is same, there is no effect on intensity 11 (d) 1 f = ( n2 n1 − 1) ( 1 R1 − 1 R2 ) where n2 and n1 are the refractive indices of the material of the lens and of the surroundings respectively. For a double concave lens, ( 1 R1 − 1 R2 ) is always negative Hence f is negative only when n2 > n1 12 (a) In given images, P,Q and R lenses in contact for P combination of lenses 1 FP = 1 f + 1 f = 2 f FP = f 2 Similarly for Q and R combination FQ = f 2 FR = f 2 Then P ∶ Q ∶ R = 1 ∶ 1 ∶ 1 13 (c) Iθ = Io cos θ = Io cos 60° = Io 2 15 (a) By using ω1 f1 + ω2 f2 = 0 ⇒ 0.02 f1 + 0.04 40 = 0 f1 = −20 cm 16 (b) 17 (b) This is a modified displacement method problem Here, a = 1.8m and a+d a−d = 2 1 d fo fe D 60° 30° 30° Incident ray Reflected ray Surface 30° n1 n1 n2 Horizon Actual position of the sun (Just below horizon) Atmospheric refraction As seen from the B The sun appears above the horizon A
Solving we get d = 0.6 m ∴ f = a 2 − d 2 4a = 0.4m 18 (c) It lamp is placed at the focus of concave mirror then we get parallel beam of light. 19 (a) P = P1 + P2 = 2D − 4D = −2D 20 (a) Our eye lens has a power to adjust its focal length to see the nearer and father objects, this process of adjusting focal length is called accommodation. However, if the object is brought too close or bring too far from the eye, the focal length cannot be adjusted to from the image on the retina. Thus, there is minimum or maximum distance for the clever vision of an object. For a normal eye, near point or least distant vision D = 25 cm and far point = ∞ 21 (c) For reading purpose u = −25 cm, v = −50 cm, f =? ∴ 1 f = 1 v − 1 u = 1 50 + 1 25 = 1 50 P = 100 f = +2D For distinct vision, f ′ =distance of far point = −3m P = 1 f′ = − 1 3 D =−0.33 D 22 (a) μ = 1.5 δm = A We know that μ = sin ( A + δm 2 ) sin A 2 1.5 = sin ( A + A 2 ) sinA 2 = sin A sin A 2 1.5 = 2 sin A 2 . cos A 2 sinA 2 1.5 = 2 cos A 2 cos A 2 = 1.5 2 = 0.75 cos41.4 = 0.75 A 2 = 41.4 A = 82.8 23 (b) Here, A = 60°, μ = √2 Now, μ = sin( A + δm 2 ) sin ( A 2 ) ... (i) Substituting given values in Eq. (i) , we get √2 = sin ( 60° + δm 2 ) sin( 60° 2 ) Or sin(30° + δm 2 ) = √2 sin30° Or sin(30° + δm 2 ) = √2 × 1 2 = 1 √2 Or sin(30° + δm 2 ) = 45° Or (30° + δm 2 ) = sin 45° Or δm = 30° ∴ Angle of incidnece i = A + δm 2 = 60° + 30° 2 = 30° 24 (b) Note that two refractive indices are involves. Therefore, two images will be formed 25 (d) 1 f = 1 v − 1 u 2.5 = 1 −0.75 − 1 u or 1 u = − 100 75 − 25 10 Or 1 u = − 4 3 − 5 2 or 1 u = −8−15 6 = − 23 6 Or u = − 6 23 m = −0.26m 26 (a) I = L r 2 27 (d) 1 f = (μ − 1) ( 1 R1 − 1 R2 ) For planoconvex lens R1 = ∞, R2 = −R = −1.5 cm, μ = 1.4 ∴ 1 f = (1.4 − 1) (0 + 1 15) or 1 f = 0.4 × 1 15 Therefore, power of the lens in diopter P = 100 f = 40 15 = 2.66 D 29 (a)
According to New Cartesian sign convention, Object distance u = −15 cm Focal length of a concave lens, f = −10 cm Height of the object ho = 2.0 cm According to mirror formula, 1 v + 1 u = 1 f 1 v = 1 f − 1 u = 1 −10 − 1 −15 ⇒ v = −30 cm This image is formed 30 cm from the mirror on the same side of the object. It is a real image Magnification of the mirror, m = −v u = hI hO ⇒ −(−30) −15 = hI 2 ⇒ hI = −4 cm Negative sign shows that image is inverted The image is real, inverted, of size 4 cm at a distance 30 cm in front of the mirror 30 (c) Focal length of effective lens 1 F = 2 fl + 1 fm = 2 fl + 1 ∞ ⇒ F = fl 2 31 (d) From graph it is clear that tan 30° = sin r sin i ⇒ 1 √3 = sinr sin i = 1 μ ⇒ μ = √3 Also v = c μ = nc ⇒ n = 1 μ = 1 √3 = (3) −1/2 32 (c) Frequency remain unchanged 33 (a) So, velocity of light in glass Vg = Vm μ Vm − Vg = Vm − Vm μ ∴ 6.25 × 107 = Vm (1 − 1 4 3 ) Vm = 6.25 × 107 × 4 = 2.5 × 108ms −1 34 (a) μblue > μred 35 (b) f = f1f2 f1 + f2 = 10(−10) 10 + (−10) = −100 10 − 10 = ∞ 36 (a) f = 1 P = 1 5 m = 20 cm Now, 1 v − 1 u = 1 20 Or 1 v − 1 −25 = 1 20 or 1 v = 1 20 − 1 25 Or 1 v = 5−4 100 or 1 v = 1 100 Or d = 100 cm = 1 m 37 (b) Sodium light gives emission spectrum having two yellow lines 38 (a) Focal length for violet is minimum 40 (a) A camera is a device used to take pictures, either singly or in sequence. Camera’s have a lens positioned in front of the camera’s opening together the incoming light and to focus the image or part of the image on the recording surface. The size of aperture (its diameter) controls the brightness of the scene control and the amount of light that enters the camera during a period of time, and the shutter controls the length of time that the light hits the recording surface. A diameter of an aperture is measured in f- stops. A lower f- stops number opens the aperture admits more light onto the camera sensor. Higher f – stop numbers make the cameras aperture smaller so less light hits the sensor. 41 (b) Only the light-gathering power is reduced 42 (d) The atmosphere can be considered to consist of a number of parallel layers of air of different densities and therefore of different refractive indices. The density and the refractive index of layers decrease with altitude. The rays of light coming from a star to the earth are thus continually refracted from the rarer to the denser layers and so they bend slightly towards the normal at each refraction from one layer to the next. Thus, they follow a curved path and reach the eyes of the observer at O as shown in figure. Hence, the image of the star S is seen as S′. But due to the wind and the convection currents in air the density of layers keep on changing and hence, the position of the stat S′ as seen, keeps on changing. These different images of the start give an impression to an observer that the star is twinkling. 43 (d) For a lens m = f−v f = − 1 f v + 1 Comparing it with y = mx + c Slope = m = − 1 f From graph, slope of the line = b c Hence − 1 f = b c ⇒ |f| = c b
44 (c) From figure, tan C = r 12 or r = 12 tan C or r = 12 sinC √1 − sin2 C r = 12 × 1 μ √1 − 1 μ 2 = 12 √μ 2 − 1 = 12 √( 4 3 ) 2 − 1 ie, r = 12 × 3 √7 45 (b) When light passes from one medium to another, its frequency remains unchanged but it velocity (and hence, wavelength) changes. aμg = wavelength in air (λa ) wavelength in glass (λg) ⇒ 1.6 = 5890 λg ⇒ λg = 5890 1.6 = 3681 Å 46 (c) fl fa = aμg − 1 lμg − 1 = aμg − 1 aμg aμl − 1 ⇒ f1 2 = 1.5 − 1 1.5 1.25 − 1 ⇒ f1 = 5cm 48 (b) v ∝ 1 μ , μ is smaller for air than water, glass and diamond 49 (c) Here, angle of incidence, i = 60° ∴ Angle of deviation δ = 180° − (i + r) = 180° − 2i (As i = r) = 180° − 2 × 60° = 60° 50 (a) Lens maker’s formula 1 f = (μ − 1) ( 1 R1 − 1 R2 ) Where, R2 = ∞, R1 = 0.3 m ∴ 1 f = ( 5 3 − 1) ( 1 0.3 − 1 ∞ ) ⇒ 1 f = 2 3 × 1 0.3 or f = 0.45 m 51 (b) Object is placed at distance 2f from the lens. So first image I will be formed at distance 2f on other side. This image I1 will behave like a virtual object for mirror. The second image I2 will be formed at distance 20 cm in front of the mirror, or at distance 10 cm to the left hand side of the lens. Now applying lens formula 1 v − 1 u = 1 f ∴ 1 v − 1 +10 = 1 +15 or v = 16 cm Therefore, the final image is at distance 16 cm from the mirror. But, this image will be real. This is because ray of light is travelling from right to left. 52 (d) From Snell’s law, refractive index (μ) is given by μ = sin i sinr ... (i) Where i is angle of incidence and r of refraction. Also, μ = v1 v2 ... (ii) Equating Eqs, (i) and (ii), we get μ = sin i sinr = v1 v2 ⇒ sin r = v2 v1 . sin i Given, v2 = 2u, v1 = u, i = 30°, sin30° = 1 2 ∴ sin r = 2 × 1 2 = 1 ⇒ r = 90° 53 (d) Critical angle for the material of prism C = sin−1 ( 1 μ ) = sin−1 ( 1 1.5 ) = 42° since angle of incidence at surface AB(60°) is greater than the critical angle (42°), so total internal reflection takes place 54 (d) ( C r c i =60o 60o A C B