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WorksheetsPROBLEM FORMULA TEST VOL II
Total questions: 29
Worksheet time: 15mins
An object is placed at a distance of 20.0 cm from a concave mirror of focal length 15.0 cm. (a) What distance from the mirror a screen should be placed to get a sharp image? (b) What is the nature of the image?
v1 +u1 = f1
v1−u1=f1
v1+u1=R1
R1+u1=f1
A thin rod of length f /3 is placed along the optical axis of a concave mirror of focal length f such that one end of image which is real and elongated just touches the respective end of the rod. Calculate the longitudinal magnification.
m1=lengthofobjectlengthofimage
m1=lengthofobjectlengthoflense
m1=lengthoflenselengthofimage
m1=lengthofimagelengthof object
Pure water has refractive index 1.33. What
is the speed of light through it?
n=vc
n=n2c
n=cv
n=cn2
Light travelling through transparent oil enters in to glass of refractive index 1.5. If the refractive index of glass with respect to the oil is 1.25, what is the refractive index of the oil?
nog=nong
ngo=nong
nog=ngno
ngo=ngno
A optical fibre is made up of a core
material with refractive index 1.68 and
a cladding material of refractive index
1.44. What is the acceptance angle of the
fibre if it is kept in air medium without
any cladding?
ia=sin−1n12−n22
ia=sin(n12−n22)
ia=sin−1n22−n12
ia=sin(n22−n12)
The thickness of a glass slab is 0.25 m.
It has a refractive index of 1.5. A ray of light
is incident on the surface of the slab at an
angle of 60o. Find the lateral displacement
of the light when it emerges from the other
side of the glass slab.
sini=nsinr
sini=sin rn
sinr=nsini
sinr=sinin
Locate the image of the point object O in the situation shown. The point C denotes the centre of curvature of the separating surface.
vn2−un1=Rn2−n1
un2−vn1=Rn2−n1
vn1−un2=Rn2−n1
vn2−un1=Rn1−n2
A biconvex lens has radii of curvature 20 cm
and 15 cm for the two curved sufaces. The
refractive index of the material of the lens
is 1.5.
(a) What is its focal length?
(b) Will the focal length change if the lens
is flipped by the side?
R1=(n−1)(R11−R21)
R1=(n−1)(R21−R11)
f1=(n−1)(R11−R21)
f1=(n−1)(R21−R11)
The wavelength of light from sodium source in vacuum is 5893Å.What are its (a) wavelength, (b) speed and (c) frequency when this light travels in water which has a
refractive index of 1.33.
λ2λ1=n2n1
λ1λ2=n2n1
λ2λ1=n1n2
λ1λ2=n1n2
Two light sources with amplitudes 5 units
and 3 units respectively interfere with each
other. Calculate the ratio of maximum and
minimum intensities.
A=a12+a22+2a1a2cosϕ
A=a12+a22−2a1a2cosϕ
A=a12+a22+2a1a2sinϕ
A=a12+a22−2a1a2sinϕ
Two light sources of equal amplitudes
interfere with each other. Calculate
the ratio of maximum and minimum
intensities.
I∝a2cos2(2ϕ)
I∝2a2cos2(2ϕ)
I∝3a2cos2(2ϕ)
I∝4a2cos2(2ϕ)
The wavelength of a light is 450 nm. How
much phase it will differ for a path of
3 mm?
λ=ϕ2π X δ
ϕ=δ2π X λ
ϕ=λ2π X δ
λ=δ2π X ϕ
In Young’s double slit experiment, the two
slits are 0.15 mm apart. The light source
has a wavelength of 450 nm. The screen is
2 m away from the slits.
(a) Find the distance of the second bright
fringe and also third dark fringe from the
central maximum.
yn=ndλD
yn=ndλD
yn=dnλD
yn=nDλd
Find the minimum thickness of a film of
refractive index 1.25, which will strongly
reflect the light of wavelength 589 nm. Also
find the minimum thickness of the film to
be anti-reflecting.
λ=d4μ
d=λ4μ
λ=4dμ
d=4μλ
Calculate the distance upto which ray
optics is a good approximation for light
of wavelength 500 nm falls on an aperture
of width 0.5 mm.
z=a2sinθ2λ
z=2λa2sinθ
z=2λa2
z=a22λ
The optical telescope in the Vainu Bappu
observatory at Kavalur has an objective
lens of diameter 2.3 m. What is its angular
resolution if the wavelength of light used is
589 nm?
θ=λa1.22
θ=1.22λa
θ=λ1.22a
θ=aλ1.22
A man with a near point of 25 cm reads
a book which has small print using a
magnifying lens of focal length 5 cm.
(a) What are the closest and the
farthest distances at which he should
keep the lens from the book? (b) What
are the maximum and the minimum
magnification possible?
u1+v1=f1
u1−v1=f1
v1+u1=f1
v1−u1=f1
A person has farsightedness with the far
distance he could see clearly is 75 cm.
Calculate the power of the lens of the
spectacles needed to rectify the defect.
f=y−25cmy x 25cm
f=y+25cmy x 25cm
f=y x 25cmy − 25cm
f=y x 25cmy + 25cm
According to wave theory, the energy
in a light wave is spread out uniformly and
continuously over the wavefront.
The energy absorbed by each electron
in time t is given by
E=tIA
E=AtI
E=IAt
E=IA
A radiation of wavelength 300 nm is incident
on a silver surface. Will photoelectrons be
observed? [work function of silver = 4.7 eV]
E=λhc(in joules)
E=λehc(in joules)
E=λhc(in eV)
E=λehc(in volts)
When light of wavelength 2200Å falls on
Cu, photo electrons are emitted from it.
Find (i) the threshold wavelength and
(ii) the stopping potential. Given: the work
function for Cu is ϕ0 = 4.65 eV.
λo=αhc
λo=θhc
λo=ϕhc
λo=γhc
Light of wavelength 390 nm is directed at
a metal electrode. To find the energy of
electrons ejected, an opposing potential
difference is established between it
and another electrode. The current of
photoelectrons from one to the other is
stopped completely when the potential
difference is 1.10 V. Determine i) the work
function of the metal
ϕ0=Kmax−hν
Kmax=ϕ0−hν
ϕ0=hν−Kmax
Kmax=ϕ0+hν
Find the de Broglie wavelength associated
with an alpha particle which is accelerated
through a potential difference of 400 V.
λ=2MqVh
λ=2eMqVh
λ=2MqVhe
λ=2MqKh
Calculate the cut-off wavelength and cutoff
frequency of x-rays from an x –ray tube
of accelerating potential 20,000 V.
λ=V124
λ=V1240
λ=V12400
λ=V X 1240
The radius of the 5th orbit of hydrogen atom is
13.25 Å. Calculate the de broglie wavelength
of the electron orbitting in the 5th orbit.
2π=nλ
2πr=λ
2πn=λ
2πr=nλ
Find the angular momentum of the electron revolving in the5th orbit of hydrogen atom.
l=2πh
l=n2πh
l=2πnh
l=2nhπ
Compute the binding energy of 2He4 nucleus using the following data: Atomic mass of Helium atom,He = 4.00260 u and that of hydrogen atom, H = 1.00785 u.
BE=ZmH+Nmn−MAc2
BE=ZmH+Nmn+MAc2
BE=[ZmH+Nmn−MA]c2
BE=[ZmH+Nmn+MA]c2
Δm=mU−mTh−mα
Δm=(mU−mTh)+mα
Δm=mU+mTh+mα
Δm=(mU−mTh−mα)c2
Calculate the number of nuclei of carbon-14 undecayed after 22,920 years if the initial number of carbon-14 atoms is 10,000. The half-life of carbon-14 is 5730 years.
n=T21t21
n=tT21
n=T21t
n=t21T21
