A bio-convex lense (refractive index 3/2) has radii of curvature 20 cm each. It is fitted to a hole in a large box filled with water (refractive index = 4/3). A point object is placed outside the box at a distance of 40 cm from lens on its axis. Distance of image formed in water is:
(a) + 160 cm
(b) + 60 cm
(c) - 100 cm
(d) + 120 cm
The following graph shows the variation of stopping potential V0 with the frequency ν of the incident radiation for two photosensitive metals X and Y:
(i) Which of the metals has larger threshold wavelength? Give reason.
(ii) Explain giving reason, which metal gives out electrons, having larger kinetic energy, for the same wavelength of the incident radiation.
(iii) If the distance between the light source and metal X is halved, how will the kinetic energy of electrons emitted from it change? Give reason
(i) Let = Frequency of incident radiations of metal Y
= Frequency of incident radiations of metal X
Therefore, metal X has larger threshold wavelength.
(ii) Since the kinetic energy of the emitted electrons is directly proportional to the frequency of incident radiation, metal Y having larger incident frequency will have larger kinetic energy.
∴ Metal Y has larger kinetic energy.
(iii) Kinetic energy of the emitted photoelectrons is independent of the intensity of the incident light. Hence, kinetic energy of the emitted photoelectrons remains unchanged if the distance between the light source and metal X is halved.
in this in (ii) part why work function.is not considered kindly reply fast... ??
Draw the circuit diagram of an illuminated photodiode in reverse bias. How is photodiode used to measure light intensity?
The circuit diagram of an illuminated photodiode in reverse bias can be represented as
The greater the intensity of light, the greater is the number of photons falling per second per unit area. Thus, the greater the intensity of light, the greater is the number of electron−hole pairs produced at the junction. The photocurrent is, thus, directly proportional to the intensity of light. This can be used for measuring the intensity of incident light.
Which of the following statements about electric field lines is incorrect?
They never cross each other
They always form closed loops
They point from positive toward negative charges
They point away from positive as well as negative charges
Electric field lines represent the electric field present in a region. The electrostatic field between a test charge and a charged body is represented by the electric field lines that exist between them. Electric field lines never cross each other and point from positive toward negative charge. They do not form closed loops and always start from a positive charge and end on a negative charge.
The correct answer is C.
Please explain how option C is incorrect ! Because I think it would be D....
State one feature by which the phenomenon of interference can be distinguished from that of diffraction.
A parallel beam of light of wavelength 600 nm is incident normally on a slit of width ‘a’. If the distance between the slits and the screen is 0.8 m and the distance of 2nd order maximum from the centre of the screen is 15 mm, calculate the width of the slit.
Difference between interference and diffraction: Interference is due to superposition of two distinct waves coming from two coherent sources. Diffraction is produced as a result of superposition of the secondary wavelets coming from different parts of the same wavefront.
Numerical: Here, λ = 600 nm = 600 × 10−19 = 6 × 10−7 m
D = 0.8 m, x = 15 mm = 1.5 × 10−3 m,
n = 2, a = ?
Name the waves that are often refered to as Heat Waves. Name the physical quantity that has a lower, higher and same valuefor these waves as compared to its value for x rays
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