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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
1001. |
The magnifying power of a simple microscope is 6. The focal length of its lens in metres will be, if least distance of distinct vision is 25cmA. `0.05`B. `0.06`C. `0.25`D. `0.12` |
Answer» Correct Answer - a | |
1002. |
Magnifying power of a simple microscope is (when final image is formed, at D=25cm from eye)A. `(D)/(f)`B. `1 + (D)/(f)`C. `1 + (f)/(D)`D. `1 -(D)/(f)` |
Answer» Correct Answer - B When final image is formed at D = 25 cm from eye. In this situation, `v=-D` from lens formula, `(1)/(v)-(1)/(u)=(1)/(f)`, we have, `(1)/(-D)-(1)/(-u)=(1)/(f)` i.e., `(D)/(u)=1+(D)/(f)` So, Magnifying power `=(D)/(u)=(1+(D)/(f))` |
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1003. |
A simple microscope is a combination of two leses of powers `+ 15 D and + 5 D` in contact. Calculate magnifying power of microscope, if final image is formed at `25 cm` from the eye. |
Answer» Here, `P = P_1 + P_2 = 15 + 5 = 20 D` Focal length of combination, `f = (100)/(P) = (100)/(20) = 5 cm` Magnifying power, `m = 1 + (d)/(f) = 1 + (25)/(5) = 6`. |
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1004. |
A ray of light falling at an angle of `50^@` is refracted through a prism and suffers minimum deviation. The angle of the prism is `60^@`. Find the angle of minimum deviation and refraction index of the material of the prism. |
Answer» Here, `i_1 = 50^@, A = 60^@, delta_m = ? , mu = ?` In position of minimum deviation, `i_2 = i_1 = 50^@` From `A + delta_m = i_1 + i_2 = 50 + 50 = 100^@` `delta_m = 100^@ - A = 100 - 60^@ = 40^@` `mu = (sin(A + delta_m)//2)/(sin A//2) = (sin(60^@ + 40^@)//2)/(sin 60^@//2) = (sin 50^@)/(sin 30^@) = (0.7660)/(0.5000) = 1.532`. |
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1005. |
Two thin equiconvex lenses each of focal length 0.2 m are placed coaxially with their optic centres 0.5m apart. Then find the focal length of the combination. |
Answer» Given, total length of each lens,`f_(1)=f_(2)=0.2m` Seperation between the lenses,d=0.5 m `therefore` Focal length of the combination, `1/F=1/(f_(1))+1/(f_(2))-d/(f_(1)f_(2))=1/0.2+1/0.2-0.5/((0.2)(0.2))=-2.5` `rArrF=-1/2.5=-0.4m` |
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1006. |
A ray of light is incident at an angle of `60^@` on the face of a prism having refracting angle `30^@.` The ray emerging out of the prism makes an angle `30^@` with the incident ray. Show that the emergent ray is perpendicular to the face through which it emerges and calculate the refractive index of the material of prism. |
Answer» Given, `i_(1)=60^(@),A=30^(@)and delta=30^(@)` From the relation, `delta=(i_(1)+i_(2))-A` we have, `i_(2)`=0 This means that the emergent ray is perpendicu,ar to the face through which it emerges. |
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1007. |
An object is placed in front of a convex mirror at a distance of 50cm. A plane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and the plane mirror is 30cm, it is found that there is no parallax between the images formed by the two mirrors. What is the radius of curvature of the convex mirror?A. `12.5 cm`B. 25 cmC. `(50)/(3) cm`D. 18 cm |
Answer» Correct Answer - b | |
1008. |
The light ray is incidence at angle of `60^(@)` on a prism of angle `45^(@)` . When the light ray falls on the other surface at `90^(@)` , the refractive index of the material of prism` mu` and the angle of devaition `delta` are given byA. `mu=sqrt(2), delta=30^(@)`B. `mu=1.5, delta=15^(@)`C. `mu=(sqrt(3))/(2), delta=30^(@)`D. `mu=sqrt((3)/(2)), delta=15^(@)` |
Answer» Correct Answer - D |
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1009. |
The light ray is incidence at angle of `60^(@)` on a prism of angle `45^(@)` . When the light ray falls on the other surface at `90^(@)` , the refractive index of the material of prism` mu` and the angle of devaition `delta` are given byA. `mu=sqrt(2),delta=30^(@)`B. `mu=1.5,delta=15^(@)`C. `mu=sqrt(3)/2,delta=30^(@)`D. `mu=sqrt(3/2),delta=15^(@)` |
Answer» Correct Answer - d | |
1010. |
A ray of light is incident at `60^(@)` on one face of a prism of angle `30^(@)` and the emergent ray makes `30^(@)` with the incident ray. The refractive index of the prism isA. 1.732B. 1.414C. 1.5D. 1.33 |
Answer» Correct Answer - A (a) As, A+ `delta`=i+e `rArr 30^(@) + 30^(@) = 60^(@)` + e Angle of emergence , e=`60^(@)-60^(@)`=0 `r_(1)=A=30^(@)` `therefore` Refractive index, `mu=(sin i)/(sin r)=sin 60^(@)/(sin 30^(@))=sqrt3=1.732` |
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1011. |
Light of wavelength is 7200 Å in air. It has a wavelength in glass `(mu=1.5)` equal toA. 7200 ÅB. 4800 ÅC. 10800 ÅD. 7201.5 Å |
Answer» Correct Answer - b | |