698 + Interview Questions in GEOMETRICAL-OPTICS IN GENERAL AWARENESS Page 13 InterviewSolution

601.	One mole of a diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperature at A,B and C are 400K, 800K and 600K respectively. Choose the correct statement: A. The change in internal energy in the process AB is `-350R`.B. The change in internal energy in the process BC is `-500R`C. The change in internal energy in whole cyclic process is `250R`.D. The change in internal energy in the process CA is `700R`.
Answer» Correct Answer - B Change in internal energy = `(fnRDeltaT)/(2) = (5)/(2)xx 1 xx R xx (-200)` =`-500R`

Discussion

602.	A glass rod has ends as shown in figure. The refractive index of glass is `mu`. The object O is at a distance 2R from the surface of larger radius of curvature. The distance between apexes of ends is 3R. Find the distance of image formed of the point object from right hand vertex. What is the condition on `mu` for formation of a real image
Answer» Correct Answer - `((9-4mu)R)/((10mu - 9)(mu-2)),2 lt mu lt 9//4`

Discussion

603.	A diver D is still under water `( mu = (4)/(3))` at a depth `d = 10 m`. A bird is diving along line AB at a constant velocity in air. When the bird is exactly above the diver he sees it at a height of 50 m from himself and velocity of the bird appears to be inclined at `45°` to the horizontal. At what distance from the diver the bird actually hits the water surface.
Answer» Correct Answer - `24.62m`

Discussion

604.	n moles of a diatomic gas has undergone a cyclic process ABC as shown in figure. Temperature at A is `T_(0)`. Find a. Volume at C ? b. Maximum temperature ? c. Total heat given to gas ? d. Is heat rejected by the gas, if yes how much heat is rejected ? e. Find out the efficiency
Answer» (i) Since triangle `OAV_(0)` and `OCA` are similar therefore `" "` `(2P)/(V)=(P_(0))/(V_(0)) implies V = 2V_(0)` Since process `AB` is isochoric hence `" "` `(P_(A))/(T_(A)) = (P_(B))/(T_(B)) implies T_(B) = 2T_(0)` Since process `BC` is isobaric therefore `" "` `(T_(B))/(V_(B)) = (T_(C))/(V_(C)) implies T_(C) = 2T_(B) = 4T_(0)` (iii) Since process is cyclic therefore `Q=W` = area under the cycle = `(1)/(2)P_(0)V_(0)` (iv) Since `U` and `W` both are negative in process `CA` `therefore Q` is negative in process `CA` and heat is rejected in process `CA` `Q_(CA) = W_(CA) + U_(CA)` =`-(1)/(2)[P_(0) + 2P_(0)]V_(0)- (5)/(2)nR(T_(c)-T_(a))` `= -(1)/(2)[P_(0) + 2P_(0)]V_(0)- (5)/(2)nR ((4P_(0)V_(0))/(nR) - (P_(0)V_(0))/(nR))` =`9P_(0)V_(0)` = Heat injected . (v) `eta = "efficieny of the cycle" = ("work done by the gas ")/("heat injected ") = (P_(0)V_(0)//2)/(Q_("injected")) xx100` where `Q_("inj") = Q_(AB) + Q_(BC) = [(5)/(2 )nR(2T_(0)-T_(0))] + [(5)/(2)nR(2T_(0)) + 2P_(0) (2V_(0) - V_(0)] = (19)/(2)P_(0)V_(0)`. Therefore `eta = (100)/(19)%`

Discussion

605.	The refractive index of light in glass varies with its wavelength according to equation `mu (lambda) = a+ (b)/(lambda^(2))` where a and b are positive constants. A nearly monochromatic parallel beam of light is incident on a thin convex lens as shown. The wavelength of incident light is `lambda_(0) +-Delta lambda` where `Delta lambda lt lt lambda_(0)`. The light gets focused on the principal axis of the lens over a region AB. If the focal length of the lens for a light of wavelength `lambda_(0)` is `f_(0)`, find the spread AB.
Answer» Correct Answer - `f_(0) [(4b Delta lambda)/(lambda_(0) (a lambda_(0)^(2) +b - lambda_(0)^(2)))]`

Discussion

606.	A ship has a green light `(lambda = 510 nm)` on its mast. What colour would be observed for this light by a diver deep inside water. Refractive index for water is `mu = (4)/(3)`.
Answer» Correct Answer - green

Discussion

607.	For a gas `(R)/(C_(P)) = 0.4`. For this gas calculate the following (i) Atomicity and degree of freedom (ii) Value of `C_(V)` and `lambda` (iii) Mean gram- molecular kinetic energy at `300K` temperature
Answer» Correct Answer - (i) mono atomic , `3` (ii) `(3)/(2)R, (5)/(3)(iii) 450R`

Discussion

608.	A gas takes part in two processes in which it is heated from the same initial state 1 to the same final temperature. The processes are shown on the p-V diagram by the straight lines 1-3 and 1-2. 2 and 3 are the points on the same isothermal curve. `Q_1` and `Q_2` are the heat transfer along the two processes. Then, A. `Q_(1) = Q_(2)`B. `Q_(1)ltQ_(2)`C. `Q_(1)gtQ_(2)`D. insufficient data
Answer» Correct Answer - B Work done in process `1-3` is greater than that in process `1-2`. While change in internal energy is same for both processes `therefore Q_(2) gtQ_(1)`.

Discussion

609.	A gas has molar heat capacity `C = 37.55 J "mole"^(-1)K^(-1)`, in the process PT = constant, find the number of degree of freedom of the molecules of the gas.
Answer» Correct Answer - `5` `PT` = constant `implies P^(2)V` = constant `implies PV^(1//2)` = constant For this process `C = C_(V) + (R)/(1-1//2) = C_(V) + 2R` `implies C_(V) = 37.35 -2 (8.314) = 20.722 = (5)/(2)R` `implies (f)/(2)R = (5)/(2)R implies f = 5`

Discussion

610.	Given T-p curve for three processes. Work done in process 1, 2 and 3 (if initial and final pressure are same for all processes) is `W_(1),W_(2)` and `W_(3)` respectively. Correct order is A. `W_(1) lt W_(2) gt W_(3)`B. `W_(1) gt W_(2) gt W_(3) `C. `W_(1) lt W_(2) lt W_(3)`D. `W_(1)=W_(2_=W_(3)`
Answer» Correct Answer - B Here `TalphaP^(n)` where for , graph-`1` , `ngt1 implies Wgt0 ,` = graph-`2` , `n=1 implies W=0`, graph-`3` `n, lt1 implies Wlt0`

Discussion

611.	One mole of an ideal gas undergoes a process whose molar heat capacity is 4R and in which work done by gas for small change in temperature is given by the relation dW=2RdT, then the ratio `(C_(P))/(C_(V))` isA. `7//5`B. `5//3`C. `3//2`D. `2`
Answer» Correct Answer - C For small change `dQ = dU+ dW` `nCdT = nC_(V)dT + 2nRdT therefore C = C_(V) + 2R, 4R = C_(V) + 2R " " therefore C_(V) = 2R` Also `C_(V) = (R)/(gamma-1) therefore (R)/(gamma-1) = 2R implies 2gamma -2 implies gamma = (3)/(2)`

Discussion

612.	A medium has `n_v = 1.56, n_r=1.44`. Then its dispersive power is:A. `3//50`B. `6//25`C. 0.03D. none of these
Answer» Correct Answer - B `omega=(n_(v)-n_(r))/(((n_(v)+n_(r))/2)-1)=6/25`

Discussion

613.	Statement I: The bulb of one thermometer is spherical while that of the other is cylindrical . Both have equal amounts of mercury. The response of the cylindrical bulb thermometer will be quicker. Statement II: Heat conduction in a body is directly proportional to cross-sectional area.A. Statement -`1` is True , Statement -`2` is True , Statement -`2` is correct explanation for Statement - `1`B. Statement -`1` is True , Statement -`2` is True , Statement -`2` is not correct explanation for Statement - `1`C. Statement -`1` is True , Statement -`2` is False.D. Statement -`1` is False , Statement -`2` is True.
Answer» Correct Answer - A

Discussion

614.	Assertion `:` The mirrorrs used in search lights are parabolic and not concave spherical. Reason `:` In a concave spherical mirrorr the image formed is always virtual.A. If both the assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If the assertion and reason both are false.
Answer» Correct Answer - C In search lights, we need an intense parallel beam of light. If a source is placed at the focus of a concave spherical mirrorr, only paraxial rays are rendered parallel. Due to large aperture of mirrorr, marginal rays give a divergent beam. But in case of parabolic mirrorr, when source is at the focus, beam of light produced over the entire cross`-` section of the mirrorr is a parallel beam.

Discussion

615.	The slit of a collimator is illuminated by a source as shown in the adjoining figures. The distance between the slit `S` and the collimating lengs `L` is equal to the focal length of the lens. The correct direction of the emergent beam will be as shown in figure. A. 1B. 3C. 2D. None of the figures.
Answer» Correct Answer - C In case of convex lens if rays are coming from the focus, then the emergent rays after refraction are parallel to principal axis.

Discussion

616.	When a calorimeter contains `40g` of water at `50^(@)C`, then the temperature falls to `45^(@)C` in `10` minutes. The same calorimeter contains `100g` of water at `50^(@)C`, it takes `20` minutes for the temperature to become `45^(@)C`. Find the water equivalent of the calorimeter.
Answer» `(m_(t)s_(t) + W)/(t_(1)) = (m_(2)s_(2) + W)/(t_(2))` where `W` is the water equivalent ` implies (40xx1+W)/(10) = (100xx1+W)/(20)` ` implies 80+2W= 100+W implies W=20g`

Discussion

617.	One gram mole of oxygen at `27^@` and one atmospheric pressure is enclosed in vessel. (i) Assuming the molecules to be moving the `V_(rms)`, Find the number of collisions per second which the molecules make with one square metre area of the vessel wall. (ii) The vessel is next thermally insulated and moved with a constant speed `V_0`. It is then suddenly stopped. The process results in a rise of the temperature of the gas by `1^@C`. Calculate the speed `V_0`.
Answer» Correct Answer - `(i) 1.96 xx 10^(27)` (ii) `36m//s` (i) Let n = number of collisions per second per unit area , change in momentum = `2`mv `therefore`Pressure exerted on wall = `overset()(n)(2mv) = P_(0)` `implies (overset()(n) xx 2 xx 32 xx 10^(-3))/(6.02 xx 10^(23)) xx sqrt((3 xx 8.3 xx 300)/(0.032))= 10^(5)` `implies overset(*)(n) = 1.95 xx 10^(27)` (ii) If vessel is suddenly stpped then `KE` will utilized in Increase in temperature. So `(1)/(2)MV_(0)^(2) = nC_(v)DeltaT` `implies V_(0) = sqrt((2nC_(v)DeltaT)/(M)) = sqrt((2C_(v)DeltaT)/(M_(w)) = 36 ms^(-1)`

Discussion

618.	Figure shows plot of `PV//T` versus `P"for" 1.00 xx 10^(-3)` kg of oxygen gas at two different temperatures. (i) What does the dotted plot signify? (ii) Which is true . `T_(1) gt T_(2) "or" T_(1) lt T_(2)`? (iii) What is the value of `PV//T` where the curves meet on the y-axis?
Answer» Correct Answer - (i) ideal gas behaviour (ii) `T_(1) gt T_(2)` (iii) 0.26JK^(-1) (i) Dotted lines correspond to ideal gas (ii) `T_(1) gt T_(2)` (On high temp. real gas behaves as ideal gas) (iii) `(PV)/(T) = (m)/(M)R = (10^(-3))/(32 xx 10^(-3)) xx 8.314 = 0.26J//K`

Discussion

619.	The work of 146 kJ is performed in order to compress one kilo mole of a gas adiabatically and in this process the temperature of the gas increases by `7^@C`. The gas is `(R=8.3ml^-1Jmol^-1K^-1)`A. diatomicB. triatomicC. A mixture of monoatomic and diatomicD. monoatomic
Answer» Correct Answer - A Work done in adiabatic process `W = (muR(T_(1)-T_(2)))/(gamma-1) implies gamma = 1 + (R(T_(2) -T_(1)))/(W)` =`1 + (10^(3) xx 8.3(7))/(146 xx 10^(3)) = 1 + 0.40 = 1.40` `therefore ` The gas must be diatomic

Discussion

620.	P-V plots for two gases during adiabatic processes are shown in the figure. Plots 1 and 2 should corresponds respectively to A. He and `O_(2)`0B. `O_(2)` and HeC. He and ArD. `O_(2) "and" N_(2)`
Answer» Correct Answer - B As `gamma_("mono") gt gamma_("dia") "so" 2 rarr "monoatomic & " 1 rarr "diatomic" `

Discussion

621.	The figure shows two paths for the change of state of a gas from A to B. The ratio of molar heat capacities in path 1 and path 2 is A. `gt1`B. `lt1`C. `1`D. Data insufficient
Answer» Correct Answer - B `Q_(1) = DeltaU + W_(1), Q_(2) = DeltaU + W_(2)` Ratio of specific heats `C_(1)/(C_(2))=(((DeltaQ_(1))/(DeltaT)))/(((DeltaQ_(2))/(DeltaT))) = (((DeltaU)/(DeltaT) + (DeltaW_(1))/(DeltaT)))/(((DeltaU)/(DeltaT) +(DeltaW_(2))/(DeltaT))) lt 1 (because W_(2)gtW_(1))`

Discussion

622.	Two identical beakers with negligible thermal expansion are filled with water to the same level at `4^(@)C`. If one says `A` is heated while the other says `B` is cooled, then:A. water level in `A` will riseB. water level in `B` will riseC. water level in `A` will fallD. water level in `B` will fall
Answer» Correct Answer - A::B Water at `4^(@)C` has highest density

Discussion

623.	Assertion: Air quickly leaking out of a balloon becomes coolers. Reason: The leaking air undergoes adiabatic expansion.A. Statement -`1` is True , Statement -`2` is True , Statement -`2` is correct explanation for Statement - `1`B. Statement -`1` is True , Statement -`2` is True , Statement -`2` is not correct explanation for Statement - `1`C. Statement -`1` is True , Statement -`2` is False.D. Statement -`1` is False , Statement -`2` is True.
Answer» Correct Answer - A

Discussion

624.	Water contained in a jar at room temperature `(20^(@)C)` is intended to be cooled by method -`I` or method `II` given below: Method -`I` : By placing ice cubes and allowing it ito float. Method-`II` : By wrapping ice cubes in a wire mesh and allowing it to sink. Choose best method(s) to cool the water.A. Method- `I` from `20^(@)C` to `4^(@)C`B. Method- `I` from `4^(@)C` to `0^(@)C`C. Method- `II` from `20^(@)C` to `4^(@)C`D. Method- `II` from `4^(@)C` to `0^(@)C`
Answer» Correct Answer - A::D Initially (above`4^(@)C`), a decreases in temperature , increases the density of water and consequently it descends , replacing the relatively warm water. Convention currents set up in this way demands the location of ice to be on the water surface. Below `4^(@)C` , a decrease in temperature decreases the water density and as a result it ascends up displacing the relatively warm water, To setup convented currents in this way , the position of ice cubes should be at the bottom.

Discussion

625.	`5`g of steam at `100^(@)C` is mixed with `10`g of ice at `0^(@)C`. Choose correct alternative /s) :- (Given `s_("water") = 1"cal"//g ^(@)C, L_(F) = 80 cal//g , L_(V) = 540cal//g`)A. Equilibrium temperature of mixture is `160^(@)C`B. Equilibrium temperature of mixture is `100^(@)C`C. At equilibrium ,mixture contains `13(1)/(3)g` of waterD. At equilibrium ,mixture contains `1(2)/(3) kg` of steam
Answer» Correct Answer - B::C::D Required heat `" "` Available heat `10`g ice (`0^(@)C`) `" "` `5` of steam (`100^(@)C`) `darr800"cal"` `" "` `darr2700"cal"` `10` g of water (`0^(@)C`) `" "` `5` g water(`100^(@)C`) `darr1000"cal"` `10` g water (`100^(@)`) So available heat is more than required heat therefore final temperature will be `100^(@)C`. Mass of vapour condensed =`(800+1000)/(540) = (10)/(3)g` Total mass of water =`10 + (10)/(3) = (40)/(3) = 13(1)/(3)`g Total mass of steam = ` 5 - (10)/(3) = (5)/(3) = 1(2)/(3)`g

Discussion

626.	A gas undergoes an adiabatic process in which pressure becomes `((8)/(3 sqrt(3)))` times and volume becomes `(3)/(4)` of initial volume. If initial absolute temperature was `T` , the final temperature isA. `(32T)/(9sqrt(3))`B. `(2T)/(sqrt(3))`C. `T^(3//2)D. `(sqrt(3)T)/(2)`
Answer» Correct Answer - B For adiabatic process `P_(1)V_(1)^(gamma) = P_(2)V_(2)^(gamma) implies P_(0)V_(0)^(gamma) = ((8)/(3sqrt(3))P_(0)) ((3)/(4)V_(0))^(gamma)` `(" …")` (i) Also , `T_(1)V_(1)^(gamma-1) = T_(2)V_(2)^(gamma-1) = implies TV_(0)^(gamma-1) = T_(2)((3)/(4)V_(0))^(gamma-1)` `( .....)` (ii) Solving (i) &(ii) , `T_(2) = (2T)/(sqrt(3))`

Discussion

627.	`5`g of steam at `100^(@)C` is mixed with `10`g of ice at `0^(@)C`. Choose correct alternative /s) :- (Given `s_("water") = 1"cal"//g ^(@)C, L_(F) = 80 cal//g , L_(V) = 540cal//g`)A. Equilibrium temperature of mixture is `160^(@)C`B. Equilibrium temperature of mixture is `100^(@)C`C. At equilibrium ,mixture contains `13(1)/(3)kg` of waterD. At equilibrium ,mixture contains `1(2)/(3) kg` of steam
Answer» Correct Answer - B::C::D Required heat `10` kg ice `(0^(@)C) " "` Available heat `5` kg steam `(100^(@)C)` `darr 800 "kcal" " " darr 2700 " Kcal"` `10` kg water `(0^(@)C) " " 5` kg water `(100 ^(@)C)` `darr 1000"kcal"` `10` kg water `(100 ^(@)C)` So available heat is more than required heat therefore final temperature will be `100^(@)C`. Mass of steam condensed =`(800+1000)/(540) = (10)/(3)`kg.Total mass of water = `10+ (10)/(3) = (40)/(3) = 13(1)/(3)kg` Total mass of steam =` 5-(10)/(3) = (5)/(3) = 1(2)/(3) kg`

Discussion

628.	The temperature drop through a two layer furnace wall is `900^@C`. Each layer is of equal area of cross section. Which of the following actions will result in lowering the temperature `theta` of the interface?A. By increasing the thermal conductivity of outer layer.B. By increasing the thermal conductivity of inner layer.C. By increasing thickness of outer layer.D. By increasing thickness of inner layer.
Answer» Correct Answer - A::D Rate of heat flow `(K_(i)A(1000 - theta))/(l_(1)) = (K_(0)A (theta - 100))/(l_(0)) implies (theta-100)/(900) = (1)/(1 + (K_(0)l_(i))/(K_(i)l_(0)))` Now, we can see that `theta` can be decrease by increasing thermal conductivity of outer layer `(K_(0))` and thickness of inner layer ` (l_(1))`.

Discussion

629.	Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are `105 N//m^2` and 6 litres respectively. The final volume of the gas are 2 litre. Molar specific heat of the gas at constant volume is `3R//2`.
Answer» Correct Answer - `-972J` `PV^(gamma)` = constant `implies (10^(5))(6)^(5//3) = (P_(2)) (2)^(5//3)` `implies P_(2) = (10^(5))(3)^(5//3) Nm^(-2)` ` W =(P_(1)V_(1) - P_(2)V_(2)) /(gamma-1)` =`((10^(5))(6 xx10^(-3)) - (10^(5) xx 3^(5//3))(2xx10^(-3)))/((5)/(3)J) = -972J`

Discussion

630.	`2m^(3)` volume of a gas at a pressure of `4xx10^(5) Nm^(-2)` is compressed adiabatically so that its volume becomes `0.5m^(3)` Find the new pressure . Compare this with the pressure that would result if the compression was isothermal. Calculate work done in each `gamma = 1.4`
Answer» `V_(1) = 2m^(3), P_(1) = 4xx10^(5)Nm^(-2)` , `" "` `V_(2) = 0.5m^(3)` In adiabatic process `" "` `P_(1)V_(1)^(gamma) = P_(2)V_(2)^(gamma) implies P_(2) = 4xx10^(5)[(2)/(0.5)]^(1.4) = 4xx10^(5) (4)^(1.4) = 2.8xx 10Nm^(-2)` In isothermal process `P_(1)V_(1) = P_(2)V_(2) implies P_(2) = (P_(1)V_(1))/(V_(2)) = (4xx10^(5)xx2)/(0.5) = 1.6 xx 10^(6)Nm^(-2).` Now work done in adiabatic process `W = (P_(1)V_(1) - P_(2)V_(2))/(gamma-1) = ((4xx10^(5) xx 2 -2.8 xx 10^(6) xx 0.5))/(1.4-1) = -1.48 xx10^(6)J` "Work done in isothermal process" `W= 2.3026RT log(V_(2))/(V_(1)) = 2.3026P_(1)V_(1)log(V_(2))/(V_(1))` ` = 2.3026 xx 4 xx 10^(5) xx 2 xx log[(0.5)/(2.0)] = 2.3026 xx 4xx 10^(5) xx 2log((1)/(4)) = -1.1 xx 10^(6)J`

Discussion

631.	Why it is cooler at the top of a mountain than at sea level?
Answer» Pressure decreases with height . Therefore if hot air rises, it suffers adiabatic expansion. From first lawof thermodynamic `Delta Q = Delta U + DeltaW implies DeltaU = - Delta W [because DeltaQ = 0]` This cause a decrease in internal energy and hence a fall of temperature.

Discussion

632.	The rainbow formed after or during the rain is due toA. total internal reflection of sunlight by rain drops suspended n airB. total internal reflection and dispersion of sunlight by the water drops suspended in airC. interference of lightD. none of these
Answer» Correct Answer - B Total internal reflection and dispersion of sunlight by the water drops suspended in air is the main reason by which rainbow is formed.

Discussion

633.	The minimum magnifying power of an astronomical telescope is M. If the focal length of its eye-lens is halved, the minimum magnifying power will become:A. M/2B. 2MC. 3MD. 4M
Answer» Correct Answer - B `MP=-((f_(0))/(f_(theta)))` If we use a eyepiece of focal length halved, then MP become double.

Discussion

634.	A luminous object and a screen are at a fixed distance D apart. A converging lens of focal length f is placed between the object and screen. A real image of the object in formed on the screen for two lens positins if they are separated by a distance d equal toA. `sqrt(D(D+4F))`B. `sqrt(D(D-4F))`C. `sqrt(2D(D-4F))`D. `sqrt(D^(2)+4F)`
Answer» Correct Answer - b. Let the objec distance be x. The, the image distance is `D-x`. From lens equation, `(1)/(x)+(1)/(D-x)=(1)/(f)` On algebraic rearrangement, we get `x^(2)-Dx+Df=0` On solving for x, we get `x_(1)=(D-sqrt(D(D-4f)))/(2)` `x_(2)=(D+sqrt(D(D-4f)))/(2)` The distance between the two object position is `d=x_(2)-x_(1)=sqrt(D(D-4f))`

Discussion

635.	A glass sphere, refractive index 1.5 and radius 10cm, has a spherical cavity of radius 5cm concentric with it. A narrow beam of parallel light is directed into the sphere. Find the final image and its nature. A. 25cm left of `S_(4)`, virtualB. 25 cm right of `S_(4)` , realC. 15 cm left of `S_(4)`, virtualD. 20cm right of `S_(4)`, virtual
Answer» Correct Answer - a. We will have single surface refractions successsively at the four surfaces `S_(1),S_(2),S_(3) and S_(4)`. Do not forget to shift origin to the vertex of respective surface. Refractive at first surface `S_(1):` Light travels from air to glass. `(1.5)/(upsilon_(1))-(1)/(oo)=((1.5-1))/((+10))` `upsilon_(1)=30cm` First image is object for the refractioni at second surface. For refraction at surface `S_(2):` Light travels from glass to air. `(1.5)/(upsilon_(2))-(1.5)/((+25))=(1-1.5)/((+5))` `upsilon_(2)=-25cm` For refraction at surface `S_(3):` Light travels from air to glass. `(1.5)/(upsilon_(3))-(1)/((-35))=((1.5-1))/((-5))` `upsilon_(3)=-35//3cm` For refraction at surface `S_(4):` Light travels from glass to air `(1)/(upsilon_(4))-(1.5)/(-(35//3+5))=(1-1.5)/(-10)` The final image is virtual, formed at 25cm to the left of the vertex of surface `S_(4)`.

Discussion

636.	A glass sphere, refractive index 1.5 and radius 10cm, has a spherical cavity of radius 5cm concentric with it. A narrow beam of parallel light is directed into the sphere. Find the final image and its nature.
Answer» We will have single surface refractions successively at the four surfaces `S_(1), S_(2),S_(3)` and `S_(4)` . Do not forget to shift origin to vertex of the respective surface. Refractive at first surface `S_(1)` : Light travel from air to glass. `(1.5)/(upsilon_(1))-(1)/(oo)=((1.5-1))/((+1))` `upsilon_(1)=30cm`. First image is the object for the rafraction at second surface. For refractive at surface `S_(2):` Light travels from glass to air. `(1)/(upsilon_(2))-(1.5)/((+25))=(1-1.5)/((+5))` `upsilon _(2)=-25 cm` For refraction at surface `S_(3):` Light travels from air to glass. `(1.5)/(upsilon_(3))-(1)/((-35))=((1.5-1))/((-5))` `upsilon _(3)=-35//3 cm` For refraction at surface `S_(4):` Light travels from glass to air. `(1)/(upsilon_(4))-(1.5)/(-(35//3+5))=(1-1.5)/(-10)` `upsilon_(4)=-25 cm` The final image is virtual, formed at 25 cm to the left of the vertex of surface `S_(4)`.

Discussion

637.	A transparent sphere of radius R ahs a cavity of radius `R//2` as shown in figure,. Find the refractive index of the sphere if a parallel beam of light falling on left sureface focuses at point P.
Answer» Correct Answer - `(3+sqrt(5))/(2)` Let refractive index of glass be `mu` Let after first refraction, image distance be v, then `(mu)/(v)-(1)/(oo)=(mu-1)/(R)rArrV=(muR)/(mu-1)` Now second refraction will take place. So distance of first image from O is `mu_(1)=(muR)/(mu-1)-R=(R)/(mu-1)` and image is formed at R. `(1)/(R) -(mu(mu-1))/(R)=(2(1-mu))/(R)rArr mu^(2)-3mu+1=0` So`mu=(3+sqrt(5))/(2)`

Discussion

638.	A microscope is focussed on a coin lying at the bottom of a beaker. The microscope is now raised up by `1 cm`. To what depth should the water be poured into the beaker so that coin is again in focus ? (Refration index of water is `4//3`)A. `1 cm`B. `4//3 cm`C. `3 cm`D. `4 cm`
Answer» Correct Answer - D Suppose water is poured up to the height `h`, So `h(1+(1)/(mu))=1impliesh=4 cm`

Discussion

639.	A lens is made of three thin different mediums. Radius of curvature and refractive index of each medium is shown iin Figure., Surface AB is straight. An object is placed at some distance from the lens by which a real image is formed on the screen placed at a distance of 10cm from the lens. Find a. The distance of the object from the lens. b. A slab of thiskness `1.5cm` and refractive index 1.5 is introduced between the image and the lens. Find the new position of the object so that image is again formed on the screen. c. Find the position of the image when same slabe is placed on the other side of the lens {as in part (b)}
Answer» Correct Answer - (a) object is at infinity. (b) 190cm (c) there will not be any change in the image formation. a. `(1)/(f_(1))=(1.6-1)[(1)/(20)-(1)/(-30)]=(1)/(20)cm^(-1)` `(1)/(f_(2))=(1.5-1)[(4)/(-30)-(1)/(15)]=(1)/(20)cm^(-1)` `(1)/(f_(3))=(2.5-1)[(1)/(15)-(1)/(oo)](1)/(10)cm^(-1)` `rArr (1)/(f_(eq)) =(1)/(20)-(1)/(20)+(1)/(10)=(1)/(10)` `rArr f_(v)=10cm` Since the image is formed on the screen which is 10 cm from the lens, so the object is at infinity. b. Shift in the image away fromt he slab due to introduction of slab of `Deltax=(1-(1)/(n))t=(1-(1)/(1.5))xx1.5=0.5` The distance of the object so tha t image is again formed on the screen `(1)/(v)-(1)/(u)=(1)/(f)` `rArr (1)/(9.5)-(1)/(u)=(1)/(10)rArr u=190cm` c. When the slab is placed on the other side of the lens, there will not be any change in the formation.

Discussion

640.	A parallel beam of light falls on the surface of a convex lens whose radius of curvature of both sides of 20cm. The refractive index of the material of the lens varies as `mu=1.5+0.5r` , where r is the distance of the point on the aperture from the optical centre in cm. Find the length of the region on the axis of the lens where the light will appear. The radius of aperture of the lens is 1cm.
Answer» Correct Answer - Light on the axis will be from 10cm to 20 cm for the lens `(1)/(f)=(mu-1)(1/R_(1)-1/R_(2))` `=(1.5+0.5r-1)(2)/(20)=0.5(1+r)(1)/(10)` `(1)/(f)=(1+f)/(20)rArrf=(20)/((1+r))` For `r=0, f=20cm`, For `r=1cm, f=10cm` Light on the axis will be from 10cm to 20cm for the lens.

Discussion

641.	A source of laser (S), a receiver (R) and a fixed mir- ror (F) – all lie on an arc of a circle of radius `R = 0.5 km`. The distance between the source and the receiver is `d = 0.5 m`. At the centre of the circle there is a small mirror M which is rotating with angular speed `omega` (see figure). Find smallest value of `omega` is if it is seen that the source shoots a laser pulse which gets reflected at M, then gets reflected at F and finally gets reflected at M to be received by the receiver.
Answer» Correct Answer - `150 rad s^(-1)`

Discussion

642.	The radii of curvature of the two surfaces of a lens are `20cm` and `30 cm` and the refractive index of the material of the lens is `1.5`. If the lens is concave`-` convex, then the focal length of lens isA. `24 cm`B. `10 cm`C. `15 cm`D. `120 cm`
Answer» Correct Answer - D `(1)/(f)=(mu-1)((1)/(R_(1))-(1)/(R_(2)))` `(1)/(f)=(1.5-1)((1)/(20)-(1)/(30))` `(1)/(f)=0.5((30-20)/(600))` or `(1)/(f)=(1)/(2)xx(10)/(600)=(1)/(20)` or `f=120cm`

Discussion

643.	An object is placed 21 cm in front of a concave mirror of radius of curvature 10cm. A glass slabe of thickness 3cm and refractive index 1.5 is then placed close to the irror in the space between the object and the mirror. The distance of the near surface of the slabe from the mirror is 1cm. The final image from the mirror will be formed atA. ``4.67cmB. 6.67cmC. 5.67cmD. 7.67cm
Answer» Correct Answer - d. `u=21cm, f=(R)/(2)=(10)/(2)=5cm` On introducing the glass slab, the object as well as the image will be shifted from the mirror through a distance `d=t(1-(1)/(mu))=3(1-(1)/(1.5))=1cm`, so that aparent distance of the object `=20cm ` i.e., `u=20cm` By the mirror formula, `(1)/(v)+(1)/(u)=(1)/(f)` `v=-(20)/(3)cm=-6.67cm` Distance of the final image from the mirror `=6.67+1` `=7.67cm`.

Discussion

644.	A converging bundle of rays is intercepted by a biconcave lens. The radii of curvature of both surfaces are 20cm and the refractive index of the material of the lens is 1.5. If the rays originially converged to a point 10cm in front of the lens, where will they now converge after passing through the lens?
Answer» Here, `mu=1.5, R_(1)=-20cm,` and `R_(2)=+20cm`. Substituting in equation, we get `f=-20cm` Now, `u=+10cm , f=-20cm` Substituting in eqution, we get `(1)/(v)=(1)/(-20)+(1)/(10)` or `v=+20cm` Thus, the rays now converge to a point 20cm in front of the lens.

Discussion

645.	A concave lens of focal length 20 cm is placed 15 cm in front of a concave mirror of radius of curvature 26 cm and further 10 cm away from the lens is placed an object. The principal axis of the lens and the mirror are coincident and the object is on the axis Find the position and nature of the image.
Answer» Correct Answer - 140 cm in front of the lens

Discussion

646.	An object AB is placed parallel and close to the optical axis between focus F and center of curvature C of a converging mirror of focale length f as shown in Figure. Then, A. Image of A will be closer than that of B from the mirror.B. Image of AB will be parallel to the optical axis.C. Image of AB will be a straight line inclined to the optical axis.D. Image of AB will not be a straight line.
Answer» Correct Answer - a.,c. The image of a point closer to the focus will be farther. As the transverse magnification of B will be more than A, the image of AB will be inclined to the optical axis.

Discussion

647.	Converging rays are incident ono a convex spherical mirror so that their extensions intersect 30cm behind the mirror on the optical axis. The reflected rays form a diverging beam so that their extensions intersect the optical axis 1.2 m from the mirror. Determine the focal length of the mirror.
Answer» In this case: `u=+30` and `v=+120` `:. (1)/(f)=(1)/(v)+(1)/(u)=(1)/(120)+(1)/(30)` `rArr f=24cm`

Discussion

648.	In Figure, find the total magnification after two successive reflections first ono `M_(1)` and then on `M_(2)`. A. `+1`B. `-2`C. `+2`D. `-1`
Answer» Correct Answer - c. For `M_(1):V=-60,m_(1)=-2` For `M_(1):u=+20,F=10` `(1)/(V)=(1)/(20)=(1)/(10)=V=20` `:.M_(2)=-(20)/(20)=-1` `:.M=m_(1)xxm_(2)=+2`

Discussion

649.	Find the position of the final image after three successive reflections taking the first reflection on `m_(1)` .
Answer» First reflection: Focus in mirror `=-10cm` `u=-15cm` Applying mirror formula: `(1)/(v)+(1)/(u)=(1)/(f)` `v=-30cm` For second reflectin on plane mirror: `u=-10cm rArr v=10cm ` For third reflection on curved mirror again: `u=-50 cm ` and `f=-10cm ` Applying mirror formula: `(1)/(v)+(1)/(u)=(1)/(f)` `rArr v=-12.5 cm`

Discussion

650.	A concave mirror gives a real image magnifies 4 times. When the object is moved 3 cm the magnification of the real image is 3 times. Find the focal length of mirror.
Answer» We know that, ` m=(v)/(u)=(f)/(u-f)` In first case, ` (f)/(u-f)=4` `F=4u-4f` and `5f=4u` In `2^(nd)` case, `(f)/((u+3)-f)=+3` `F=3u+9-3f` `:. 4u-4f=3u+9-3f` `u=f+9=(4u)/(5)+9` `u=45 cm` and `f=36 cm`

Discussion

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