31 and 32 1 68% 7:41 PMLTE26. A convex spherical mirror has a radius of curvature of 60 cm and an object is placed 40 cm in frorof the mirror. Find the image distance.a. -17.14 cmb. 17.14 cmс. 120 cmd. --120 cm27. Which one is corect as far as the image?a. larger, inverted and real.c. smaller, upright and real.e. smaller, inverted and virtual.b. larger, upright and virtuald. smaller, upright and virtual.28. The index of refraction of crown glass is 1.523 and the index of refraction of alcohol is 1.362. Ligenters from glass to alcohol, find the critical angle when total internal reflection occurs.a. 63.4 degreesb. 41.0 degreesc. 47.2 degreesd. 75.6 degreesA diverging lens has a focal length of - 40 cm. A virtual image is formed at a distance of 30 cm fromthe lens, on the same side as the object.(Hint: you have to take q = -30 cm because of the virtuaimage.)29. Find the object distance.а. - 120 cmb. -17.14 cmc. 17.14 cmd. 120 cm30. find the magnification of the lens.a. 0.25b. -0.25c. 4.0d. 1.75A very long non-conducting wire with radius R is charged uniformly with volume charge density of p31. Use the Gauss Law to find the electric field inside the wire.(rR)toa. pr/3ɛ,c. pR?/2€,re. 4pR/3ɛ,rb. pr/2ɛ,d. pR/3ɛ,rΣ

Gauss law The net electric flux through any hypothetical closed surface is equal to 1 ε 0 {\displaystyle {\frac {1}{\varepsilon _{0}}}} times the net electric charge within that closed surface. ϕE=Qε031. r<RConsider a Gaussian surface of length 'l' electric field at point p at distance 'r' from central axis total charge enclosed by Gaussian surface qenc=ρπr2l By Guass law for surface ∮E·ds =qencε0=ρπx2lε0E∫ds =ρπx2lε0E×2πrl=ρπr2lε0E=ρr2ε0Electric field inside long wire at distance r from central axis Option 'b' is correct 32. r>R Consider Gaussian surface, we will find electric field at point P ∮E·ds =qencε0=ρπR2lε0E∫ds =ρπR2lε0E×(2πrl)=ρπR2lε0E=ρR22ε0rElectric field outside of wire charge with unifom volume charge desinity ρ option 'c' is correct.