Calculate the radius of curvature of an equi-concave lens of refractive index 1.5, when it is kept in a medium of refractive index 1.4, to have a power of -5D?
CBSE 55-1-1 PAPER SET 2019
Download our app for free and get started
Using lens maker formulae for given equi-cancave lens,
$\frac{1}{\text{f}}=\Big(\frac{\text{n}_2}{\text{n}_1}-1\Big)\Big(\frac{1}{-\text{R}}-\frac{1}{\text{R}}\Big)$
$\Rightarrow\frac{1}{\text{f}}=\Big(\frac{\text{n}_2}{\text{n}_1}-1\Big)\Big(-\frac{2}{\text{R}}\Big)\ ...(1)$
Where $n_2$ and $n_1 n_2$ and $n_1$ are the refractive index of the lens and respectively. where $n_2=1.5 \& n_1=1.4$
Power of lens = -5D (Given)
Focal length, $\text{f}=\frac{1}{-5}\times100=-20\text{cm}$
Putting all the values in equation 1, we get,
$\Rightarrow\frac{1}{-20}=\Big(\frac{1.5}{1.4}-1\Big)\Big(-\frac{2}{\text{R}}\Big)$
$\Rightarrow\frac{1}{20}=\Big(\frac{0.1}{1.4}\Big)\Big(\frac{2}{\text{R}}\Big)$
$\Rightarrow\text{R}=\frac{4}{1.4}=2.86\text{cm}$
$\frac{1}{\text{f}}=\Big(\frac{\text{n}_2}{\text{n}_1}-1\Big)\Big(\frac{1}{-\text{R}}-\frac{1}{\text{R}}\Big)$
$\Rightarrow\frac{1}{\text{f}}=\Big(\frac{\text{n}_2}{\text{n}_1}-1\Big)\Big(-\frac{2}{\text{R}}\Big)\ ...(1)$
Where $n_2$ and $n_1 n_2$ and $n_1$ are the refractive index of the lens and respectively. where $n_2=1.5 \& n_1=1.4$
Power of lens = -5D (Given)
Focal length, $\text{f}=\frac{1}{-5}\times100=-20\text{cm}$
Putting all the values in equation 1, we get,
$\Rightarrow\frac{1}{-20}=\Big(\frac{1.5}{1.4}-1\Big)\Big(-\frac{2}{\text{R}}\Big)$
$\Rightarrow\frac{1}{20}=\Big(\frac{0.1}{1.4}\Big)\Big(\frac{2}{\text{R}}\Big)$
$\Rightarrow\text{R}=\frac{4}{1.4}=2.86\text{cm}$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1View SolutionLight of wavelength 560nm goes through a pinhole of diameter 0.20mm and falls on a wall at a distance of 2.00m. What will be the radius of the central bright spot formed on the wall?
- 2View SolutionDraw the output waveform at X, using the given inputs A and B for the logic circuit shown below. Also, identify the logic operation performed by this circuit.
- 3View SolutionThe refractive index of a material is. What is the angle of refraction if the unpolarised light is incident on it at the polarising angle of the medium?
- 4Find the angular separation between the consecutive bright fringes in a Young's double slit experiment with blue-green light of wavelength 500nm. The separation between the slits is $2.0 × 10^{-3}$m.View Solution
- 5View SolutionIs it necessary to have two waves of equal intensity to study interference pattern? Will there be an effect on clarity if the waves have unequal intensity?
- 6Define the term 'Half-life' of a radioactive substance. Two different radioactive substances have half-lives $T_1$ and $T_2$ and number of undecayed atoms at an instant $N_1$ and $N_2$, respectively. Find the ratio of their activities at that instant.View Solution
- 7View SolutionA convex lens of diameter 8.0cm is used to focus a parallel beam of light of wavelength 620nm. If the light be focused at a distance of 20cm from the lens, what would be the radius of the central bright spot formed?
- 8View SolutionA soap film of thickness 0.0011mm appears dark when seen by the reflected light of wavelength 580nm. What is the index of refraction of the soap solution, if it is known to be between 1.2 and 1.5?
- 9View SolutionIn double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1º. What is the spacing between the two slits?
- 10View SolutionNo interference pattern is detected when two coherent sources are infinitely close to each other. Why?