Question

^{A laser emitted a 0.1 ms long pulse of light with total energy 10 J. The average pressure of such a pulse focused on a spot of diameter 1mm on a fully reflective surface is: OA. 850 Pa B. 0.17 mN OC. 6.6 N/m2 D. None of the above}

Answer 1

& Given energy of light O.Im sec . The Power E = 105 in emitted by laser of loses p = energy, time P = 10J J 0.1 X10 sec 10 10 105 Watt Pulse diameter of The Radius will be m The Pressure extesed pl = 2I O=1mm R = 0.5mm 0.5 5x103 by light on fully reflective surface I is intensity of light I = Power los Area (= 3x108 m/sec where 2 TR? p'- 2x105 TR²C 8

pa 2X105 á (0.5x16%)* x3x108 0.84883X10 10? 848.83 850 Poscop 850 Pa. The consect n=5.5 (Air ņ21) Pressure N/m² Pressure extered by Pulse is option is Q. Given refractive index let say goes from germentum to air at angle of genmenium incidence O, .n,=55 using snells low n, siro, = n, sino Aire Where Qc (angle of Caiticay then O. nosiro e (1) sin(4) = h e = sin o incidence) 2 2 h, sin Oc =1 siä" (55) Sin Oc --- - 10.48 5.5 So incidence or critical angle is The angle of critical Og = 10.489 conx C+ Option PS D The doo = 4em Giden distance focal length of of object from lens i converging lens ! = + ioa to cm )

+ +1 using this fi dol using this lens formula ㅗ ㅗ I for dol doi lens formula de ㅗ fidon (1014 din doi doi-fi 4-10 (dili 40 =-6.67 cm (ave sign implies that image -6. will form same side of imageld) will form at distance object) from lens 1 (dil, - - 6.67 cm ter ² dpi as ... - I object at - di 20cm 1 el OS I (final image) K of imege (II) is The magnification - Sopla mos -66-67) dol m =+ 1.67 ( tue sign implies that image will be upright) Now the image Is acts the object for lens 2 distance which virtual object doa 6.67 +20 doz=26.67 cm lens 2 from = 10 cm using formula f doz dia fa dos (10) (26.67) ~ 16 em doo - f 26.67-10 plored at face Lt 1 dia

The image 16cm fo -(16) -0.6 dor 26.67 m-0.6 will be formed at distance. die from lens 2 -dia The magnification of lens 2 Me The (-ve sign implies that image I will be inverted The final mognification m = mi ma = (+1.67) (-0.6) m =-1.0 The magnitude of m=1 (so image will be some size as object) and image image will Prverted. 1.01 3 be