A convex lens made of glass `(mu_(g)=3//2)` has focal length f in air. The image of an object placed in front of it is inverted, real and magnified. Now the whole arrangement distance between object and lens. ThenA. the new focal length will becom 4fB. the new focal length will become `f//4`C. new image will be virtual and magnifiedD. new image will be real, inverted and smaller in size

A convex lens made of glass `(mu_(g)=3//2)` has focal length f in air.

1.	A convex lens made of glass `(mu_(g)=3//2)` has focal length f in air. The image of an object placed in front of it is inverted, real and magnified. Now the whole arrangement distance between object and lens. ThenA. the new focal length will becom 4fB. the new focal length will become `f//4`C. new image will be virtual and magnifiedD. new image will be real, inverted and smaller in size
Answer» Correct Answer - a.,c. `(1)/(f_(air))=((3)/(2)-1)((1)/(R_(1))-(1)/(R_(2)))` `(1)/(f_(water))=((3//2)/(4//3)-1)((1)/(R_(1))-(1)/(R_(2)))` From these two equations we get, `f_(water)=4f_(air)=4f` In air image was inverted, real and magnified. Therefore, object was lying between f and 2f. Now the focal length has changed to 4f. Therefore, the object now lies between pole and focus. Hence, the new image will be virtual and magnified.

Discussion

No Comment Found

Related InterviewSolutions

A symmetric doule convex lens is cut in two equal parts by a plane perpendiculr to the pricipal axis. If the power of the original lens was 4D, the power of a cut lens will beA. 2DB. 3DC. 4DD. 5D
Three perfect gases at absolute temperature `T_(1), T_(2)` and `T_(3)` are mixed. The masses f molecules are `m_(1), m_(2)` and `m_(3)` and the number of molecules are `n_(1), n_(2)` and `n_(3)` respectively. Assuming no loss of energy, the final temperature of the mixture isA. `(n_(1)T_(1)^(2) + n_(2)T_(2)^(2) + n_(3)T_(3)^(2))/(n_(1)T_(1) + n_(2)T_(2) + n_(3)T_(3))`B. `(n_(1)^(2)T_(1)^(2) + n_(2)^(2)T_(2)^(2) + n_(3)^(2)T_(3)^(2))/(n_(1)T_(1) + n_(2)T_(2) + n_(3)T_(3))`C. `(T_(1) + T_(2) + T_(3))/(3)`D. `(n_(1)T_(1) + n_(2)T_(2) + n_(3)T_(3))/(n_(1) + n_(2) + n_(3))`
The critical angle for water is `48.2^@`. Find is refractive index.
Find the size of the image formed in the situation shown in figure.
The angle of minimum, deviation from a prism is `37^@`. If the angle of prism is `53^@`, find the refrence index of the material of the prism.
A prism of refractive index `1.53` is placed in water of refractive index `1.33`. If the angle of prism is `60^@`, calculate the angle of minimum deviation in water.
Light passes symmetrically through a `60^(@)` prism of refractive index 1.54. After emergence out from the prism the light ray is incident on a plane mirror fixed to the base of the prism extending beyond it. Find the total deviation of the light ray after reflection from the mirror.
A vertical beam of light of cross sectional radius `(R)/(2)` is incident symmetrically on the curved surface of a glass hemisphere of refractive index `mu = (3)/(2)`. Radius of the hemisphere is R and its base is on a horizontal table. Find the radius of luminous spot formed on the table. `sin 20^(@) = (1)/(3)` and `sin 80^(@) = 0.98`
An obsever looks at a distant tree of height 10m with a telescope of magnifying power of 20. to the observer the tree appears:A. 20 times nearerB. 10 times tallerC. 10 times nearerD. 20 times taller
A large transparent cube (refractive index = 1.5) has a small air bubble inside it. When a coin (diameter 2 cm) is placed symmetrically above the bubble on the top surface of the cube, the bubble cannot be seen by looking down into the cube at any angle. However, when a smaller coin (diameter 1.5 cm) is placed directly over it, the bubble can be seen by looking down into the cube. What is the range of the possible depths d of the air bubble beneath the top surface ?

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment