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PROBLEM FORMULA TEST VOL II

Total questions: 29

Worksheet time: 15mins

Name
Class
Date
1.

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?

a)

1v +1u = 1f \frac{1}{v}\ +\frac{1}{u}\ =\ \frac{1}{f}\ \

b)

1v1u=1f\frac{1}{v}-\frac{1}{u}=\frac{1}{f}

c)

1v+1u=1R\frac{1}{v}+\frac{1}{u}=\frac{1}{R}

d)

1R+1u=1f\frac{1}{R}+\frac{1}{u}=\frac{1}{f}

2.

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.

a)

m1=lengthofimagelengthofobjectm_1=\frac{lengthofimage}{lengthofobject}

b)

m1=lengthoflenselengthofobjectm_1=\frac{lengthoflense}{lengthofobject}

c)

m1=lengthofimagelengthoflensem_1=\frac{lengthofimage}{lengthoflense}

d)

m1=lengthof objectlengthofimagem_1=\frac{lengthof\ object}{lengthofimage}

3.

Pure water has refractive index 1.33. What

is the speed of light through it?

a)

n=cvn=\frac{c}{v}

b)

n=cn2n=\frac{c}{n_2}

c)

n=vcn=\frac{v}{c}

d)

n=n2cn=\frac{n_2}{c}

4.

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?

a)

nog=ngnon_{og}=\frac{n_g}{n_o}

b)

ngo=ngnon_{go}=\frac{n_g}{n_o}

c)

nog=nongn_{og}=\frac{n_o}{n_g}

d)

ngo=nongn_{go}=\frac{n_o}{n_g}

5.

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?

a)

ia=sin1n12n22i_a=\sin^{-1}\sqrt{n_1^2-n_2^2}

b)

ia=sin(n12n22)i_a=\sin\left(\sqrt{n_1^2-n_2^2}\right)

c)

ia=sin1n22n12i_a=\sin^{-1}\sqrt{n_2^2-n_1^2}

d)

ia=sin(n22n12)i_a=\sin\left(\sqrt{n_2^2-n_1^2}\right)

6.

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.

a)

sini=sinrn\sin i=\frac{\sin r}{n}

b)

sini=nsin r\sin i=\frac{n}{\sin\ r}

c)

sinr=sinin\sin r=\frac{\sin i}{n}

d)

sinr=nsini\sin r=\frac{n}{\sin i}

7.

Locate the image of the point object O in the situation shown. The point C denotes the centre of curvature of the separating surface.

a)

n2vn1u=n2n1R\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}

b)

n2un1v=n2n1R\frac{n_2}{u}-\frac{n_1}{v}=\frac{n_2-n_1}{R}

c)

n1vn2u=n2n1R\frac{n_1}{v}-\frac{n_2}{u}=\frac{n_2-n_1}{R}

d)

n2vn1u=n1n2R\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_1-n_2}{R}

8.

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?

a)

1R=(n1)(1R11R2)\frac{1}{R}=\left(n-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

b)

1R=(n1)(1R21R1)\frac{1}{R}=\left(n-1\right)\left(\frac{1}{R_2}-\frac{1}{R_1}\right)

c)

1f=(n1)(1R11R2)\frac{1}{f}=\left(n-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

d)

1f=(n1)(1R21R1)\frac{1}{f}=\left(n-1\right)\left(\frac{1}{R_2}-\frac{1}{R_1}\right)

9.

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.

a)

λ1λ2=n1n2\frac{\lambda_1}{\lambda_2}=\frac{n_1}{n_2}

b)

λ2λ1=n1n2\frac{\lambda_2}{\lambda_1}=\frac{n_1}{n_2}

c)

λ1λ2=n2n1\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}

d)

λ2λ1=n2n1\frac{\lambda_2}{\lambda_1}=\frac{n_2}{n_1}

10.

Two light sources with amplitudes 5 units

and 3 units respectively interfere with each

other. Calculate the ratio of maximum and

minimum intensities.

a)

A=a12+a22+2a1a2cosϕA=\sqrt{a_1^2+a_2^2+2a_1a_2\cos\phi}

b)

A=a12+a222a1a2cosϕA=\sqrt{a_1^2+a_2^2-2a_1a_2\cos\phi}

c)

A=a12+a22+2a1a2sinϕA=\sqrt{a_1^2+a_2^2+2a_1a_2\sin\phi}

d)

A=a12+a222a1a2sinϕA=\sqrt{a_1^2+a_2^2-2a_1a_2\sin\phi}

11.

Two light sources of equal amplitudes

interfere with each other. Calculate

the ratio of maximum and minimum

intensities.

a)

Ia2cos2(ϕ2)I\propto a^2\cos^2\left(\frac{\phi}{2}\right)

b)

I2a2cos2(ϕ2)I\propto2a^2\cos^2\left(\frac{\phi}{2}\right)

c)

I3a2cos2(ϕ2)I\propto3a^2\cos^2\left(\frac{\phi}{2}\right)

d)

I4a2cos2(ϕ2)I\propto4a^2\cos^2\left(\frac{\phi}{2}\right)

12.

The wavelength of a light is 450 nm. How

much phase it will differ for a path of

3 mm?

a)

λ=2πϕ X δ\lambda=\frac{2\pi}{\phi}\ X\ \delta

b)

ϕ=2πδ X λ\phi=\frac{2\pi}{\delta}\ X\ \lambda

c)

ϕ=2πλ X δ\phi=\frac{2\pi}{\lambda}\ X\ \delta

d)

λ=2πδ X ϕ\lambda=\frac{2\pi}{\delta}\ X\ \phi

13.

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.

a)

yn=λDndy_n=\frac{\lambda D}{nd}

b)

yn=nλDdy_n=n\frac{\lambda D}{d}

c)

yn=dλDny_n=d\frac{\lambda D}{n}

d)

yn=nλdDy_n=n\frac{\lambda d}{D}

14.

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.

a)

λ=4μd\lambda=\frac{4\mu}{d}

b)

d=4μλd=\frac{4\mu}{\lambda}

c)

λ=μ4d\lambda=\frac{\mu}{4d}

d)

d=λ4μd=\frac{\lambda}{4\mu}

15.

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.

a)

z=2λa2sinθz=\frac{2\lambda}{a^2\sin\theta}

b)

z=a2sinθ2λz=\frac{a^2\sin\theta}{2\lambda}

c)

z=a22λz=\frac{a^2}{2\lambda}

d)

z=2λa2z=\frac{2\lambda}{a^2}

16.

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?

a)

θ=1.22λa\theta=\frac{1.22}{\lambda a}

b)

θ=λa1.22\theta=\frac{\lambda a}{1.22}

c)

θ=1.22aλ\theta=\frac{1.22a}{\lambda}

d)

θ=a1.22λ\theta=a\frac{1.22}{\lambda}

17.

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?

a)

1u+1v=1f\frac{1}{u}+\frac{1}{v}=\frac{1}{f}

b)

1u1v=1f\frac{1}{u}-\frac{1}{v}=\frac{1}{f}

c)

1v+1u=1f\frac{1}{v}+\frac{1}{u}=\frac{1}{f}

d)

1v1u=1f\frac{1}{v}-\frac{1}{u}=\frac{1}{f}

18.

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.

a)

f=y x 25cmy25cmf=\frac{y\ x\ 25cm}{y-25cm}

b)

f=y x 25cmy+25cmf=\frac{y\ x\ 25cm}{y+25cm}

c)

f=y 25cmy x 25cmf=\frac{y\ -\ 25cm}{y\ x\ 25cm}

d)

f=y + 25cmy x 25cmf=\frac{y\ +\ 25cm}{y\ x\ 25cm}

19.

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

a)

E=IAtE=\frac{IA}{t}

b)

E=IAtE=\frac{I}{At}

c)

E=IAt

d)

E=IA

20.

A radiation of wavelength 300 nm is incident

on a silver surface. Will photoelectrons be

observed? [work function of silver = 4.7 eV]

a)

E=hcλ(in joules)E=\frac{hc}{\lambda}\left(in\ joules\right)

b)

E=hcλe(in joules)E=\frac{hc}{\lambda e}\left(in\ joules\right)

c)

E=hcλ(in eV)E=\frac{hc}{\lambda}\left(in\ eV\right)

d)

E=hcλe(in volts)E=\frac{hc}{\lambda e}\left(in\ volts\right)

21.

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.

a)

λo=hcα\lambda_o=\frac{hc}{\alpha}

b)

λo=hcθ\lambda_o=\frac{hc}{\theta}

c)

λo=hcϕ\lambda_o=\frac{hc}{\phi}

d)

λo=hcγ\lambda_o=\frac{hc}{\gamma}

22.

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

a)

ϕ0=Kmaxhν\phi_0=K_{\max}-h\nu

b)

Kmax=ϕ0hνK_{\max}=\phi_0-h\nu

c)

ϕ0=hνKmax\phi_0=h\nu-K_{\max}

d)

Kmax=ϕ0+hνK_{\max}=\phi_0+h\nu

23.

Find the de Broglie wavelength associated

with an alpha particle which is accelerated

through a potential difference of 400 V.

a)

λ=h2MqV\lambda=\frac{h}{\sqrt{2M_qV}}

b)

λ=h2eMqV\lambda=\frac{h}{\sqrt{2eM_qV}}

c)

λ=he2MqV\lambda=\frac{he}{\sqrt{2M_qV}}

d)

λ=h2MqK\lambda=\frac{h}{\sqrt{2M_qK}}

24.

Calculate the cut-off wavelength and cutoff

frequency of x-rays from an x –ray tube

of accelerating potential 20,000 V.

a)

λ=124V\lambda=\frac{124}{V}

b)

λ=1240V\lambda=\frac{1240}{V}

c)

λ=12400V\lambda=\frac{12400}{V}

d)

λ=V X 1240λ=V\ X\ 1240​

25.

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.

a)

2π=nλ2\pi=n\lambda

b)

2πr=λ2\pi r=\lambda

c)

2πn=λ2\pi n=\lambda

d)

2πr=nλ2\pi r=n\lambda

26.

Find the angular momentum of the electron revolving in the5th orbit of hydrogen atom.

a)

l=h2πl=\frac{h}{2\pi}

b)

l=hn2πl=\frac{h}{n2\pi}

c)

l=nh2πl=\frac{nh}{2\pi}

d)

l=hπ2nl=\frac{h\pi}{2n}

27.

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.

a)

BE=ZmH+NmnMAc2BE=Zm_H+Nm_n-M_Ac^2

b)

BE=ZmH+Nmn+MAc2BE=Zm_H+Nm_n+M_Ac^2

c)

BE=[ZmH+NmnMA]c2BE=\left[Zm_H+Nm_n-M_A\right]c^2

d)

BE=[ZmH+Nmn+MA]c2BE=\left[Zm_H+Nm_n+M_A\right]c^2

28.
a)

Δm=mUmThmα\Delta m=m_U-m_{Th}-m_{\alpha}

b)

Δm=(mUmTh)+mα\Delta m=\left(m_U-m_{Th}\right)+m_{\alpha}

c)

Δm=mU+mTh+mα\Delta m=m_U+m_{Th}+m_{\alpha}

d)

Δm=(mUmThmα)c2\Delta m=\left(m_U-m_{Th}-m_{\alpha}\right)c^2

29.

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.

a)

n=t12T12n=\frac{t_{\frac{1}{2}}}{T_{\frac{1}{2}}}

b)

n=T12tn=\frac{T_{\frac{1}{2}}}{t}

c)

n=tT12n=\frac{t}{T_{\frac{1}{2}}}

d)

n=T12t12n=\frac{T_{\frac{1}{2}}}{t_{\frac{1}{2}}}